Download - Aat Solutions - Ch12

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.

667Algebra 2

Worked-Out Solution Key

Prerequisite Skills (p. 792)

1. The domain of f (x) is all real numbers except x Þ 0.

2. The range of g(x) is all real numbers.

3. The composition f (g(x)) is equal to 1 }

4x 1 2 .

4. 7x 1 3 5 31 5. 9 5 2x 2 7

7x 5 28 16 5 2x

x 5 4 8 5 x

6. 14 5 23x 1 8 7. 10 2 3x 5 28

6 5 23x 23x 5 18

22 5 x x 5 26

8. 11x 1 9 5 3x 1 17 9. 2x 1 3 5 26 2 x

8x 5 8 3x 5 29

x 5 1 x 5 23

10. 3x 1 y 5 0 3 4 12x 1 4y 5 0

22x 2 4y 5 230 22x 2 4y 5 230

10x 5 230

x 5 23

3(23) 1 y 5 0 → y 5 9

The solution is (23, 9).

11. 2x 2 2y 5 10 2x 2 2y 5 10

x 1 y 5 210 3 2 2x 1 2y 5 220

4x 5 210

x 5 2 5 } 2

2 5 } 2 1 y 5 210 → y 5 2

15 } 2

The solution is 1 2 5 } 2 , 2

15 } 2 2 .

12. 4x 2 5y 5 25 4x 2 5y 5 25

0.5x 1 1.5y 5 18.5 3 28 24x 2 12y 5 2148

217y 5 2123

y 5 123

} 17

4x 2 5 1 123 }

17 2 5 25 → x 5

260 } 17

The solution is 1 260 }

17 ,

123 }

17 2 .

13. f (g(x)) 5 2(22x21) 2 1 5 24x21 2 1

Domain: all real numbers except x 5 0

14. f ( f (x)) 5 2(2x 2 1) 2 1 5 4x 2 3

Domain: all real numbers

15. g(g(x)) 5 22(22x21)21 5 x

Domain: all real numbers except x 5 0

Lesson 12.1

12.1 Guided Practice (pp. 794–797)

1. a1 5 1 1 4 5 5 2. f (1) 5 (22)1 2 1 5 1

a2 5 2 1 4 5 6 f (2) 5 (22)2 2 1 5 22

a3 5 3 1 4 5 7 f (3) 5 (22)3 2 1 5 4

a4 5 4 1 4 5 8 f (4) 5 (22)4 2 1 5 28

a5 5 5 1 4 5 9 f (5) 5 (22)5 2 1 5 16

a6 5 6 1 4 5 10 f (6) 5 (22)6 2 1 5 232

3. a1 5 1 } 1 1 1 5

1 } 2

a2 5 2 } 2 1 1 5

2 } 3

a3 5 3 } 3 1 1 5

3 } 4

a4 5 4 } 4 1 1 5

4 } 5

a5 5 5 } 5 1 1 5

5 } 6

a6 5 6 } 6 1 1 5

6 } 7

4. The sequence 3, 8, 15, 24, . . .

5

n21

an

can be written as 1 p 3, 2 p 4, 3 p 5, 4 p 6, . . .. The next term is a5 5 5 p 7 5 35. A rule for the nth term is an 5 n(n 1 2).

5. a9 5 92 5 81

There are 81 apples in the 9th layer.

6. ai 5 5i

Lower limit 5 1

Upper limit 5 20

Summation notation: i 5 1

∑ 20

5i

7. ai 5 i2

} i2 1 1

Lower limit 5 1

Upper limit 5 infi nity


∑ `

i2 }

i2 1 1

8. ai 5 6i

Lower limit 5 1



∑ `

6i

9. 4 1 i

Lower limit 5 1

Upper limit 5 8


∑ 8

(4 1 i)

Chapter 12

n2ws-1200-a.indd 667 6/27/06 11:30:51 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.

668Algebra 2Worked-Out Solution Key

10. i 5 1

∑ 5

8i 5 8(1) 1 8(2) 1 8(3) 1 8(4) 1 8(5)

5 8 1 16 1 24 1 32 1 40

5 120

11. k 5 3

∑ 7

(k2 2 1) 5 (32 2 1) 1 (42 2 1) 1 (52 2 1) 1

(62 2 1) 1 (72 2 1) 5 8 1 15 1 24 1 35 1 48

5 130

12. i 5 1

∑ 34

1 5 34

13. n 5 1

∑ 6

n 5 1 1 2 1 3 1 4 1 5 1 6 5 21

14. n 5 1

∑ 9

n2 5 12 1 22 1 32 1 42 1 52 1 62 1 72 1 82 1 92

5 1 1 4 1 9 1 16 1 25 1 36 1 49 1 64 1 81

5 285

There are 285 apples in the stack.

12.1 Exercises (pp. 798–800)

Skill Practice

1. Another name for summation notation is sigma notation.

2. A sequence is a list of numbers and a series is the sum of the terms in a sequence.

3. a1 5 1 1 2 5 3 4. a1 5 6 2 1 5 5

a2 5 2 1 2 5 4 a2 5 6 2 2 5 4

a3 5 3 1 2 5 5 a3 5 6 2 3 5 3

a4 5 4 1 2 5 6 a4 5 6 2 4 5 2

a5 5 5 1 2 5 7 a5 5 6 2 5 5 1

a6 5 6 1 2 5 8 a6 5 6 2 6 5 0

5. a1 5 12 5 1 6. f (1) 5 13 1 2 5 3

a2 5 22 5 4 f (2) 5 23 1 2 5 10

a3 5 32 5 9 f (3) 5 33 1 2 5 29

a4 5 42 5 16 f (4) 5 43 1 2 5 66

a5 5 52 5 25 f (5) 5 53 1 2 5 127

a6 5 62 5 36 f (6) 5 63 1 2 5 218

7. a1 5 41 2 1 5 1 8. a1 5 212 5 21

a2 5 42 2 1 5 4 a2 5 222 5 24

a3 5 43 2 1 5 16 a3 5 232 5 29

a4 5 44 2 1 5 64 a4 5 242 5 216

a5 5 45 2 1 5 216 a5 5 252 5 225

a6 5 46 2 1 5 1024 a6 5 262 5 236

9. f (1) 5 12 2 5 5 24 10. a1 5 (1 1 3)2 5 16

f (2) 5 22 2 5 5 21 a2 5 (2 1 3)2 5 25

f (3) 5 32 2 5 5 4 a3 5 (3 1 3)2 5 36

f (4) 5 42 2 5 5 11 a4 5 (4 1 3)2 5 49

f (5) 5 52 2 5 5 20 a5 5 (5 1 3)2 5 64

f (6) 5 62 2 5 5 31 a6 5 (6 1 3)2 5 81

11. f (1) 5 2 4 } 1 5 24 12. a1 5

3 } 1 5 3

f (2) 5 2 4 } 2 5 22 a2 5

3 } 2

f (3) 5 2 4 } 3 a3 5

3 } 3 5 1

f (4) 5 2 4 } 4 5 21 a4 5

3 } 4

f (5) 5 2 4 } 5 a5 5

3 } 5

f (6) 5 2 4 } 6 5 2

2 } 3 a6 5

3 } 6 5

1 } 2

13. a1 5 2(1)

} 1 1 2 5 2 } 3 14. f (1) 5

1 } 2(1) 2 1 5 1

a2 5 2(2)

} 2 1 2 5 1 f (2) 5 2 } 2(2) 2 1 5

2 } 3

a3 5 2(3)

} 3 1 2 5 6 } 5 f (3) 5

3 } 2(3) 2 1 5

3 } 5

a4 5 2(4)

} 4 1 2 5 4 } 3 f (4) 5

4 } 2(4) 2 1 5

4 } 7

a5 5 2(5)

} 5 1 2 5 10

} 7 f (5) 5 5 } 2(5) 2 1 5

5 } 9

a6 5 2(6)

} 6 1 2 5 3 } 2 f (6) 5

6 } 2(6) 2 1 5

6 } 11

15. Given terms: 5 p 1 2 4, 5 p 2 2 4, 5 p 3 2 4, 5 p 4 2 4, . . .

Next term: 5 p 5 2 4 5 21

Rule for nth term: 5n 2 4

16. Given terms: 21 2 1, 22 2 1, 23 2 1, 24 2 1, . . .

Next term: 25 2 1 5 16


17. Given terms: (21)1(4 p 1), (21)2(4 p 2), (21)3(4 p 3), (1)4(4 p 4), . . .

Next term: (21)5(4 p 5) 5 220

Rule for nth term: (21)n(4n)

18. Given terms: 13 1 1, 23 1 1, 33 1 1, 43 1 1, . . .

Next term: 53 1 1 5 126

Rule for nth term: n3 1 1

19. Given terms: 2 }

3 p 1 , 2 }

3 p 2 , 2 }

3 p 3 , 2 }

3 p 4 , . . .

Next term: 2 }

3 p 5 5 2 } 15

Rule for nth term: 2 }

3n

20. Given terms: 1 p 2

} 1 1 2

, 2 p 2

} 2 1 2

, 3 p 2

} 3 1 2

, 4 p 2

} 4 1 2

, . . .

Next term: 5 p 2

} 5 1 2

5 10

} 7

Rule for nth term: 2n }

2 1 n

21. Given terms: 1 }

4 ,

2 }

4 ,

3 }

4 ,

4 }

4 ,

5 }

4 , . . .

Next term: 6 }

4

Rule for nth term: n }

4

Chapter 12, continued

n2ws-1200-a.indd 668 6/28/06 1:48:37 PM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.

669Algebra 2



22. Given terms: 2 p 1 2 1

} 1 p 10

, 2 p 2 2 1

} 2 p 10

, 2 p 3 2 1

} 3 p 10

, 2 p 4 2 1

} 4 p 10

, . . .

Next term: 2 p 5 2 1

} 5 p 10

5 9 } 50


} 10n

23. Given terms: 2.4 1 0.7(1), 2.4 1 0.7(2), 2.4 1 0.7(3), 2.4 1 0.7(4), . . .

Next term: 2.4 1 0.7(5) 5 5.9

Rule for nth term: 2.4 1 0.7n

24. Given terms: 5.8 2 1.6(1), 5.8 2 1.6(2), 5.8 2 1.6(3), 5.8 2 1.6(4), 5.8 2 1.6(5), . . .

Next term: 5.8 2 1.6(6) 5 23.8


25. Given terms: 0.2 1 12, 0.2 1 22, 0.2 1 32, 0.2 1 42, . . .

Next term: 0.2 1 52 5 25.2

Rule for nth term: 0.2 1 n2

26. Given terms: 1.2 1 7.8(1), 1.2 1 7.8(2), 1.2 1 7.8(3), 1.2 1 7.8(4), . . .

Next term: 1.2 1 7.8(5) 5 40.2


27. D;

a1 5 1(1 1 1)

} 2 5 1

a2 5 2(2 1 1)

} 2 5 3

a3 5 3(3 1 1)

} 2 5 6

a4 5 4(4 1 1)

} 2 5 10

28.2

n21

an 29.

n1

7

an

30.

3

n21

an

31. an

4

n21

32.

4

n21

an 33.

4

n21

an

34.

5

n21

an 35.

1

n21

an

36.

1

n21

an

37. ai 5 3i 1 4

Lower limit 5 1

Upper limit 5 5


∑ 5

(3i 1 4)

38. 6i 2 1

Lower limit 5 1

Upper limit 5 5


∑ 5

(6i 2 1)

39. 2i 2 3

Lower limit 5 1



∑ `

(2i 2 3)

n2ws-1200-a.indd 669 6/28/06 1:49:13 PM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.


40. (22)i

Lower limit 5 1



∑ `

(22)i

41. 7i 2 4

Lower limit 5 1



∑ `

(7i 2 4)

42. 1 }

3i

Lower limit 5 1

Upper limit 5 4


∑ 4

1 }

3i

43. i }

3 1 i

Lower limit 5 1

Upper limit 5 7


∑ 7

i }

3 1 i

44. i2 2 2

Lower limit 5 1



∑ `

(i2 2 2)

45. i 5 1

∑ 6

2i 5 2 p 1 1 2 p 2 1 2 p 3 1 2 p 4 1 2 p 5 1 2 p 6

5 2 1 4 1 6 1 8 1 10 1 12

5 42

46. i 5 1

∑ 5

7i 5 7 p 1 1 7 p 2 1 7 p 3 1 7 p 4 1 7 p 5

5 7 1 14 1 21 1 28 1 35

5 105

47. n 5 0

∑ 4

n3 5 03 1 13 1 23 1 33 1 43

5 0 1 1 1 8 1 27 1 64

5 100

48. k 5 1

∑ 4

3k2 5 3 p 12 1 3 p 22 1 3 p 32 1 3 p 42

5 3 1 12 1 27 1 48

5 90

49. k 5 3

∑ 6

(5k 2 2) 5 (5 p 3 2 2) 1 (5 p 4 2 2)

1 (5 p 5 2 2) 1 (5 p 6 2 2)

5 13 1 18 1 23 1 28

5 82

50. n 5 1

∑ 5

(n2 2 1) 5 (12 2 1) 1 (22 2 1) 1 (32 2 1) 1 (42 2 1) 1 (52 2 1) 5 0 1 3 1 8 1 15 1 24

5 50

51. i 5 1

∑ 8

2 } i 5

2 } 1 1

2 } 2 1

2 } 3 1

2 } 4 1

2 } 5 1

2 } 6 1

2 } 7 1

2 } 8

5 2 1 1 1 2 } 3 1

1 } 3 1

1 } 2 1

1 } 4 1

2 } 5 1

2 } 7

5 761

} 140

52. k 5 1

∑ 6

k }

k 1 1 5

1 } 1 1 1 1

2 } 2 1 1 1

3 } 3 1 1

1 4 } 4 1 1 1

5 } 5 1 1 1

6 } 6 1 1

5 1 } 2 1

2 } 3 1

3 } 4 1

4 } 5 1

5 } 6 1

6 } 7

5 617

} 140

53. i 5 1

∑ 35

1 5 35

54. n 5 1

∑ 16

n 5 1 1 2 1 3 1 4 1 5 1 6 1 . . . 1 14 1 15 1 16

5 136

55. i 5 1

∑ 25

i 5 1 1 2 1 3 1 4 1 5 1 . . . 1 23 1 24 1 25

5 325

56. n 5 1

∑ 18

n2 5 12 1 22 1 32 1 42

1 52 1 . . . 1 162 1 172 1 182

5 1 1 4 1 9 1 16

1 25 1 . . . 1 256 1 289 1 324

5 2109

57. The fi rst term is missing. The lower limit is 0, so the fi rst term should be 2(0) 1 3, or 3. Correct sum: 3 1 5 1 7 1 9 1 11 1 13 5 48

58. B;

i 5 1

∑ 20

i 5 1 1 2 1 3 1 4 1 . . . 1 17 1 18 1 19 1 20

5 210

59. true;

i 5 1

∑ n

kai 5 ka1 1 ka2 1 ka3 1 ka4 1 . . . 1 kan

5 k(a1 1 a2 1 a3 1 . . . 1 an)

5 k i 5 1

∑ n

ai

60. true;

i 5 1

∑ n

(ai 1 bi) 5 (a1 1 b1) 1 (a2 1 b2)

1 (a3 1 b3) 1 . . . 1 (an 1 bn)

5 (a1 1 a2 1 a3 1 . . . 1 an)

1 (b1 1 b2 1 b3 1 . . . 1 bn)

5 i 5 1

∑ n

ai 1 i 5 1

∑ n

bi


n2ws-1200-a.indd 670 6/27/06 11:31:12 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.

671Algebra 2


61. false;

Sample answer: Let ai 5 2i and bi 5 24i.

i 5 1

∑ 3

(2i)(24i) 5 (2 p 1)(24 p 1) 1 (2 p 2)(24 p 2)

1 (3 p 2)(24 p 3)

5 28 1 (232) 1 (272)

5 2112

i 5 1

∑ 3

2i 5 2 p 1 1 2 p 2 1 2 p 3 5 12

i 5 1

∑ 3

(24i) 5 24 p 1 1 (24 p 2) 1 (24 p 3) 5 224

1 i 5 1

∑ 3

2i 2 1 i 5 1

∑ 3

(24i) 2 5 12(224) 5 2288

Because 2112 Þ 2288, i 5 1

∑ n

aibi Þ 1 i 5 1

∑ n

ai 2 1 i 5 1

∑ n

bi 2 . 62. false;

Sample answer: Let ai 5 i and k 5 2.

i 5 1

∑ 3

i2 5 12 1 22 1 32 5 1 1 4 1 9 5 14

1 i 5 1

∑ 3

i 2 2

5 (1 1 2 1 3)2 5 62 5 36

Because 14 Þ 36, i 5 1

∑ n

aik Þ 1

i 5 1 ∑

n

ai 2 k.Problem Solving

63. a3 5 180(3 2 2)

} 3 5 608

a4 5 180(4 2 2)

} 4 5 908

a5 5 180(5 2 2)

} 5 5 1088

a6 5 180(6 2 2)

} 6 5 1208

a7 5 180(7 2 2)

} 7 ø 128.578

Tn 5 180(n 2 2)

} n p n 5 180(n 2 2)

Total measure of angles in skylight 5 180(12 2 2), or 18008.

64. Use Special Series Formula for sum of fi rst n positive

integers: i 5 1

∑ n

i 5 n(n 1 1)

} 2

In 100 days, you will have 100(100 1 1)

}} 2 5 5050 pennies,

or $50.50.

To save $500, you need 50,000 pennies.

n(n 1 1)

} 2 5 50,000

n(n 1 1) 5 100,000

n2 1 n 2 100,000 5 0

(n 2 Ï}

100,000 )2 5 0

n 5 Ï}

100,000

5 316.23

You need about 316 days to save $500.

65. Given terms: 21 2 1, 22 2 1, 23 2 1, 24 2 1, 25 2 1

Rule for nth term: an 5 2n 2 1

a6 5 26 2 1 5 63

a7 5 27 2 1 5 127

a8 5 28 2 1 5 255

You need 63 moves to move 6 rings, 127 moves to move 7 rings, and 255 moves to move 8 rings.

66. a. d4 5 0.3(2)4 2 2 1 0.4 5 1.6

Mars is 1.6 astronomical units from the sun.

b. 149,600,000 km

}} 1 a.u.

3 1.6 a.u. 5 239,360,000 km

Mars is about 239,360,000 kilometers from the sun.

c.

Position of planet from sun

Mea

n d

ista

nce

fro

m s

un

(a.

u.)

0 1 2 3 4 5 6 7 8 9 n

dn

048

12162024283236

67. a. a5 5 5(5 1 1)

} 2 5 15

There are 15 balls in the fi fth layer.

b. a1 1 a2 1 a3 1 a4 1 a5

5 1(1 1 1)

} 2 1 2(2 1 1)

} 2 1 3(3 1 1)

} 2 1 4(4 1 1)

} 2 1 5(5 1 1)

} 2

5 1 1 3 1 6 1 10 1 15

5 35

There are 35 balls in the stack.

c. The difference in each layer is

n2 2 n(n 1 1)

} 2 5 2n2 2 n2 2 n

}} 2

5 n2 2 n

} 2

5 n(n 2 1)

} 2 .

Each layer of the square pyramid contains exactly

n(n 2 1)

} 2 more balls than the corresponding layer in the

triangular pyramid.

68. The number of balls in the top n layers is

i 5 1

∑ n

n(n 1 1)

} 2 5

i 5 1 ∑

n

n2 1 n

} 2 .

Using Exercise 59: i 5 1

∑ n

n2 1 n

} 2 5

1 } 2 i 5 1

∑ n

(n2 1 n)

Using Exercise 60: 1 }

2 i 5 1

∑ n

(n2 1 n) 5 1 } 2 1

i 5 1 ∑

n

n2 1 i 5 1

∑ n

n 2 Using the special formulas on page 797:

1 }

2 1

i 5 1 ∑

n

n2 1 i 5 1

∑ n

n 2 5 1 } 2 1 n(n 1 1)(2n 1 1)

}} 6 1

n(n 1 1) } 2 2


n2ws-1200-a.indd 671 6/27/06 11:31:17 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.


Mixed Review

69. 18 5 4x 2 2 70. 23 5 6x 2 1

20 5 4x 24 5 6x

5 5 x 4 5 x

71. 12 5 2 2 5x

10 5 25x

22 5 x

72. 5 5 8x 2 5 73. 17 5 20 2 2x

10 5 8x 23 5 22x

5 }

4 5 x

3 }

2 5 x

74. 14 5 5x 1 4 75. 7 2 3x 5 16

10 5 5x 23x 5 9

2 5 x x 5 23

76. 9 1 8x 5 25 77. 11x 2 6 5 239

8x 5 16 11x 5 233

x 5 2 x 5 23

78. d 5 Ï}}

(6 2 4)2 1 (1 2 (25))2

5 Ï}

4 1 36

5 2 Ï}

10

79. d 5 Ï}}}

(22 2 (27))2 1 (21 2 4)2

5 Ï}

25 1 25

5 5 Ï}

2

80. d 5 Ï}}

(5 2 0)2 1 (22 2 5)2

5 Ï}

25 1 49

5 Ï}

74

81. d 5 Ï}}

(1 2 (24))2 1 (9 2 6)2

5 Ï}

25 1 9

5 Ï}

34

82. d 5 Ï}}}

(6 2 2)2 1 (24 2 (25))2

5 Ï}

16 1 1

5 Ï}

17

83. d 5 Ï}}}

(22 2 (25))2 1 (28 2 (24))2

5 Ï}

9 1 16

5 5

84. d 5 Ï}}

(5 2 9)2 1 (6 2 7)2

5 Ï}

16 1 1

5 Ï}

17

85. d 5 Ï}}

(3 2 (21))2 1 (2 2 8)2

5 Ï}

16 1 36

5 2 Ï}

13

86. d 5 Ï}}

(29 2 4)2 1 (26 2 0)2

5 Ï}

169 1 36

5 Ï}

205

Graphing Calculator Activity 12.1 (p. 801)

1. a. 5, 9, 13, 17, 21, 25, 29, 33, 37, 41

b.

c. Sum 5 230

2. a. 9, 12, 15, 18, 21, 24, 27, 30, 33, 36

b.

c. Sum 5 225

3. a. 32, 29, 26, 23, 20, 17, 14, 11, 8, 5

b.

c. Sum 5 185

4. a. 17, 19, 21, 23, 25, 27, 29, 31, 33, 35

b.

c. Sum 5 260

5. a. 4, 7, 12, 19, 28, 39, 52, 67, 84, 103

b.

c. Sum 5 415

6. a. 1, 2, 4, 8, 16, 32, 64, 128, 256, 512

b.

c. Sum 5 1023


n2ws-1200-a.indd 672 6/27/06 11:31:27 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.

673Algebra 2


Lesson 12.2


1. a2 2 a1 5 14 2 17 5 23

a3 2 a2 5 11 2 14 5 23

a4 2 a3 5 8 2 11 5 23

a5 2 a4 5 5 2 8 5 23

Each difference is 23, so the sequence is arithmetic.

2. a1 5 17; d 5 14 2 17 5 23

an 5 17 1 (n 2 1)(23) 5 20 2 3n

a20 5 20 2 3(20) 5 240

3. a11 5 a1 1 (11 2 1) d

257 5 a1 1 10(27)

13 5 a1

an 5 13 1 (n 2 1)(27) 5 20 2 7n

a20 5 20 2 7(20) 5 2120

4. a7 5 26 5 a1 1 (7 2 1)d → 26 5 a1 1 6d

a16 5 71 5 a1 1 (16 2 1)d → 71 5 a1 1 15d

45 5 9d

5 5 d

71 5 a1 1 15(5) → a1 5 24

an 5 24 1 (n 2 1)(5) 5 29 1 5n

a20 5 29 1 5(20) 5 91

5. a1 5 2 1 7(1) 5 9

a12 5 2 1 7(12) 5 86

S12 5 12 1 9 1 86 }

2 2 5 570

6. a1 5 3(1) 5 3

a8 5 3(8) 5 24

S8 5 8 1 3 1 24 }

2 2 5 108

The house of cards contains 108 cards.

12.2 Exercises (pp. 806–809)

Skill Practice

1. The constant difference between consecutive terms of an arithmetic sequence is called the common difference.

2. An arithmetic sequence is a list of numbers that have the same common difference between consecutive terms. An arithmetic series is the sum of the terms.

3. a2 2 a1 5 22 2 1 5 23

a3 2 a2 5 25 2 (22) 5 23

a4 2 a3 5 28 2 (25) 5 23

a5 2 a4 5 211 2 (28) 5 23

The sequence is arithmetic with common difference 23.

4. a2 2 a1 5 14 2 16 5 22

a3 2 a2 5 11 2 14 5 23

The differences are not constant, so the sequence is not arithmetic.

5. a2 2 a1 5 14 2 5 5 9

a3 2 a2 5 23 2 14 5 9

a4 2 a3 5 32 2 23 5 9

a5 2 a4 5 41 2 32 5 9

The sequence is arithmetic with common difference 9.

6. a2 2 a1 5 27 2 (210) 5 3

a3 2 a2 5 25 2 (27) 5 2


7. a2 2 a1 5 1 2 0.5 5 0.5

a3 2 a2 5 1.5 2 1 5 0.5

a4 2 a3 5 2 2 1.5 5 0.5

a5 2 a4 5 2.5 2 2 5 0.5

The sequence is arithmetic with common difference 0.5.

8. a2 2 a1 5 10 2 20 5 210

a3 2 a2 5 5 2 10 5 25


9. a2 2 a1 5 5 } 4 2

7 } 4 5 2

2 } 4

a3 2 a2 5 3 } 4 2

5 } 4 5 2

2 } 4

a4 2 a3 5 2 3 } 4 2

3 } 4 5 2

6 } 4


10. a2 2 a1 5 2 } 7 2

1 } 7 5

1 } 7

a3 2 a2 5 4 } 7 2

2 } 7 5

2 } 7


11. a2 2 a1 5 21 2 1 2 5 } 2 2 5

3 } 2

a3 2 a2 5 1 } 2 2 (21) 5

3 } 2

a4 2 a3 5 2 2 1 } 2 5

3 } 2

a5 2 a4 5 7 } 2 2 2 5

3 } 2

The sequence is arithmetic with common difference 3 }

2 .

12. a1 5 1; d 5 4 2 1 5 3

an 5 1 1 (n 2 1)(3) 5 3n 2 2

a20 5 3(20) 2 2 5 58

13. a1 5 5; d 5 11 2 5 5 6

an 5 5 1 (n 2 1)(6) 5 6n 2 1

a20 5 6(20) 2 1 5 119

14. a1 5 8; d 5 21 2 8 5 13

an 5 8 1 (n 2 1)(13) 5 13n 2 5

a20 5 13(20) 2 5 5 255


n2ws-1200-a.indd 673 6/27/06 11:31:31 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.


15. a1 5 23; d 5 21 2 (23) 5 2

an 5 23 1 (n 2 1)(2) 5 2n 2 5

a20 5 2(20) 2 5 5 35

16. a1 5 6; d 5 2 2 6 5 24

an 5 6 1 (n 2 1)(24) 5 24n 1 10

a20 5 24(20) 1 10 5 270

17. a1 5 25; d 5 14 2 25 5 211

an 5 25 1 (n 2 1)(211) 5 211n 1 36

a20 5 211(20) 1 36 5 2184

18. a1 5 0; d 5 2 } 3 2 0 5

2 } 3

an 5 0 1 (n 2 1) 1 2 } 3 2 5

2 } 3 n 2

2 } 3

a20 5 2 } 3 (20) 2

2 } 3 5

38 } 3

19. a1 5 2; d 5 2 5 } 3 2 2 5 2

1 } 3

an 5 2 1 (n 2 1) 1 2 1 } 3 2 5 2

1 } 3 n 1

7 } 3

a20 5 2 1 } 3 (20) 1

7 } 3 5 2

13 } 3

20. a1 5 1.5; d 5 3.6 2 1.5 5 2.1

an 5 1.5 1 (n 2 1)(2.1) 5 2.1n 2 0.6

a20 5 2.1(20) 2 0.6 5 41.4

21. The general rule for an arithmetic sequence is an 5 a1 1 (n 2 1)d, not an 5 a1 1 nd. So,

an 5 37 1 (n 2 1)(213) 5 213n 1 50.

22. The values of 37 and 213 were substituted in the wrong place. It should be an 5 37 1 (n 2 1)(213), so an 5 213n 1 50.

23. a16 5 a1 1 (16 2 1)d 3

n1

an

52 5 a1 1 15(5)

223 5 a1

an 5 223 1 (n 2 1)(5)

an 5 5n 2 28

n 1 2 3 4 5 6

an 223 218 213 28 23 2

24. a6 5 a1 1 (6 2 1)d 8

n

an

21 216 5 a1 1 5(9)

261 5 a1

an 5 261 1 (n 2 1)(9)

an 5 9n 2 70

n 1 2 3 4 5 6

an 261 252 243 234 225 216

25. a4 5 a1 1 (4 2 1)d

14

n1

an

96 5 a1 1 3(214)

138 5 a1

an 5 138 1 (n 2 1)(214)

an 5 214n 1 152

n 1 2 3 4 5 6

an 138 124 110 96 82 68

26. a12 5 a1 1 (12 2 1)d

8

n21

an

23 5 a1 1 11(27)

74 5 a1

an 5 74 1 (n 2 1)(27)

an 5 27n 1 81

n 1 2 3 4 5 6

an 74 67 60 53 46 39

27. a10 5 a1 1 (10 2 1)d

2

n21

an

30 5 a1 1 9 1 7 } 2 2

2 3 } 2 5 a1

an 5 2 3 } 2 1 (n 2 1) 1 7 }

2 2

an 5 7 } 2 n 2 5

n 1 2 3 4 5 6

an 21.5 2 5.5 9 12.5 16


n2ws-1200-a.indd 674 6/27/06 11:31:41 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.

675Algebra 2


28. a11 5 a1 1 (11 2 1)d

1

n21

an

1 }

2 5 a1 1 10 1 2

1 } 2 2

11

} 2 5 a1

an 5 11

} 2 1 (n 2 1) 1 2 1 } 2 2

an 5 6 2 1 } 2 n

n 1 2 3 4 5 6

an 5.5 5 4.5 4 3.5 3

29. C;

a30 5 a1 1 (30 2 1)d

57 5 a1 1 29(4)

259 5 a1

an 5 259 1 (n 2 1)(4)

an 5 263 2 4n

30. a10 5 a1 1 (10 2 1)d → 85 5 a1 1 9d

a4 5 a1 1 (4 2 1)d → 31 5 a1 1 3d

54 5 6d

9 5 d

85 5 a1 1 9(9) → a1 5 4

an 5 4 1 (n 2 1)(9) 5 25 1 9n

31. a14 5 a1 1 (14 2 1)d → 79 5 a1 1 13d

a6 5 a1 1 (6 2 1)d → 39 5 a1 1 5d

40 5 8d

5 5 d

79 5 a1 1 13(5) → a1 5 14

an 5 14 1 (n 2 1)(5) 5 9 1 5n

32. a17 5 a1 1 (17 2 1)d → 40 5 a1 1 16d

a3 5 a1 1 (3 2 1)d → 22 5 a1 1 2d

42 5 14d

3 5 d

40 5 a1 1 16(3) → a1 5 28

an 5 28 1 (n 2 1)(3) 5 211 1 3n

33. a20 5 a1 1 (20 2 1)d → 258 5 a1 1 19d

a8 5 a1 1 (8 2 1)d → 210 5 a1 1 7d

248 5 12d

24 5 d

258 5 a1 1 19(24) → a1 5 18

an 5 18 1 (n 2 1)(24) 5 22 2 4n

34. a15 5 a1 1 (15 2 1)d → 137 5 a1 1 14d

a9 5 a1 1 (9 2 1)d → 89 5 a1 1 8d

48 5 6d

8 5 d

137 5 a1 1 14(8) → a1 5 25

an 5 25 1 (n 2 1)(8) 5 17 1 8n

35. a11 5 a1 1 (11 2 1)d → 35 5 a1 1 10d

a2 5 a1 1 (2 2 1)d → 17 5 a1 1 d

18 5 9d

2 5 d

35 5 a1 1 10(2) → a1 5 15

an 5 15 1 (n 2 1)(2) 5 13 1 2n

36. a12 5 a1 1 (12 2 1)d → 29 5 a1 1 11d

a7 5 a1 1 (7 2 1)d → 4 5 a1 1 6d

213 5 5d

2 13

} 5 5 d

29 5 a1 1 11 1 2 13

} 5 2 → a1 5 98

} 5

an 5 98

} 5 1 (n 2 1) 1 2 13

} 5 2 5 111

} 5 2 13

} 5 n

37. a9 5 a1 1 (9 2 1)d → 24 5 a1 1 8d

a5 5 a1 1 (5 2 1)d → 15 5 a1 1 4d

9 5 4d

9 }

4 5 d

24 5 a1 1 8 1 9 } 4 2 → a1 5 6

an 5 6 1 (n 2 1) 1 9 } 4 2 5

15 } 4 2

9 } 4 n

38. a11 5 a1 1 (11 2 1)d → 22 5 a1 1 10d

a6 5 a1 1 (6 2 1)d → 0 5 a1 1 5d

22 5 5d

2 2 } 5 5 d

22 5 a1 1 10 1 2 2 } 5 2 → a1 5 2

an 5 2 1 (n 2 1) 1 2 2 } 5 2 5

12 } 5 2

2 } 5 n

39. B;

a13 5 a1 1 (13 2 1)d → 248 5 a1 1 12d

a6 5 a1 1 (6 2 1)d → 26 5 a1 1 5d

242 5 7d

26 5 d

248 5 a1 1 12(26) → a1 5 24

an 5 24 1 (n 2 1)(26) 5 30 2 6n

40. a1 5 1 1 3(1) 5 4

a10 5 1 1 3(10) 5 31

S10 5 10 1 4 1 31 }

2 2 5 175

41. a1 5 23 2 2(1) 5 25

a8 5 23 2 2(8) 5 219

S8 5 8 1 25 1 (219) }

2 2 5 296

42. a1 5 14 2 6(1) 5 8

a18 5 14 2 6(18) 5 294

S18 5 18 1 8 1 (294) }

2 2 5 2774


n2ws-1200-a.indd 675 6/27/06 11:31:46 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.


43. a1 5 29 1 11(1) 5 2

a22 5 29 1 11(22) 5 233

S22 5 22 1 2 1 233 }

2 2 5 2585

44. a3 5 72 2 6(3) 5 54

a9 5 72 2 6(9) 5 18

S7 5 7 1 54 1 18 }

2 2 5 252

45. a5 5 254 1 9(5) 5 29

a14 5 254 1 9(14) 5 72

S10 5 10 1 29 1 72 }

2 2 5 315

46. a1 5 2

a15 5 58

S15 5 15 1 2 1 58 }

2 2 5 450

47. a1 5 21

a8 5 34

S8 5 8 1 21 1 34 }

2 2 5 132

48. a1 5 44

a7 5 2

S7 5 7 1 2 1 44 }

2 2 5 161

49. 2, 7, 12, 17, . . .

a1 5 2, d 5 5

an 5 2 1 (n 2 1)(5) 5 23 1 5n

50. 21, 24, 27, 210, . . .

a1 5 21, d 5 23

an 5 21 1 (n 2 1)(23) 5 2 2 3n

51. 23, 25, 27, 29, . . .

a1 5 23, d 5 22

an 5 23 1 (n 2 1)(22) 5 21 2 2n

52. The graph of an is a scatter plot whose points lie on the line that represents the graph of f (x).

53. false; Sample answer: Consider the series 2 1 4 1 6 1 8, whose sum is 20. If d is doubled, the series becomes 2 1 6 1 10 1 14, whose sum is 32. Because 32 Þ 2(20), the statement is false.

54. true;

Because an is an arithmetic sequence, the fi rst three terms a, b, and c, are a, a 1 n, and a 1 2n.

b 0 1 }

2 (a 1 c)

a 1 n 0 1 }

2 (a 1 (a 1 2n))

a 1 n 0 1 }

2 (2a 1 2n)

a 1 n 5 a 1 n ✓

55. i 5 1

∑ n

(25 1 7i) 5 486

n1 a1 1 an }

2 2 5 486

na1 1 nan 5 486(2)

n(2) 1 n(25 1 7n) 5 972

7n2 2 3n 2 972 5 0

(7n 1 81)(n 2 12) 5 0

n 5 2 81

} 7 or n 5 12

The number of terms must be positive, so n 5 12.

56. i 5 1

∑ n

(10 2 3i) 5 228

n1 a1 1 an }

2 2 5 228

na1 1 nan 5 228(2)

n(7) 1 n(10 2 3n) 5 256

23n2 1 17n 1 56 5 0

(23n 2 7)(n 2 8) 5 0

n 5 8 or n 5 2 7 } 3


57. i 5 1

∑ n

(58 2 8i) 5 21150

n1 a1 1 an }

2 2 5 21150

na1 1 nan 5 21150(2)

n(50) 1 n(58 2 8n) 5 22300

28n2 1 108n 1 2300 5 0

24(n 2 25)(2n 1 23) 5 0

n 5 25 or n 5 2 23

} 2


58. i 5 1

∑ n

(5 2 5i) 5 250

n1 a1 1 an }

2 2 5 250

na1 1 nan 5 250(2)

n(0) 1 n(5 2 5n) 5 2100

25n2 1 5n 1 100 5 0

25(n 2 5)(n 1 4) 5 0

n 5 5 or n 5 24



n2ws-1200-a.indd 676 6/27/06 11:31:50 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.

677Algebra 2


59. i 5 3

∑ n

(23 2 4i) 5 2507

(n 2 2) 1 a3 1 an }

2 2 5 2507

(n 2 2)(a3 1 an) 5 2507(2)

(n 2 2)[215 1 (23 2 4n)] 5 21014

(n 2 2)(218 2 4n) 5 21014

218n 2 4n2 1 36 1 8n 5 21014

24n2 2 10n 1 1050 5 0

22(n 2 15)(2n 1 35) 5 0

n 5 15 or n 5 217.5


60. i 5 5

∑ n

(7 1 12i) 5 455

(n 2 4) 1 a5 1 an }

2 2 5 455

(n 2 4)(a5 1 an) 5 455(2)

(n 2 4)[67 1 (7 1 12n)] 5 910

(n 2 4)(74 1 12n) 5 910

74n 1 12n2 2 296 2 48n 5 910

12n2 1 26n 2 1206 5 0

2(6n 1 67)(n 2 9) 5 0

n 5 9 or n 5 2 67

} 6


61. i 5 1

∑ 150

(2i 2 1) 5 n 1 a1 1 a150 }

2 2

5 150 1 1 1 299 }

2 2

5 22,500

62. S3 5 n 1 a1 1 a3 }

2 2

3 2 x 1 x 1 1 2 3x 5 3 1 3 2 x 1 1 2 3x }}

2 2

23x 1 4 5 3 1 24x 1 4 } 2 2

23x 1 4 5 26x 1 6

3x 5 2

x 5 2 } 3

a1 5 3 2 2 } 3 5

7 } 3

a2 5 2 } 3

a3 5 1 2 3 1 2 } 3 2 5 21

Because the common difference is 2 5 } 3 , the next term is

21 2 5 } 3 , or 2

8 } 3 .

Problem Solving

63. a. a1 5 6, d 5 6

an 5 6 1 (n 2 1)(6)

an 5 6n

b. Total number of cells 5 S9 1 1

S9 5 9 1 a1 1 a9 }

2 2

S9 5 9 1 6 1 54 }

2 2

S9 5 270

There are 271 cells in the honeycomb.

64. a1 5 3, d 5 2 S7 5 7 1 a1 1 a7 }

2 2

an 5 3 1 (n 2 1)(2) S7 5 7 1 3 1 15 }

2 2

an 5 1 1 2n S7 5 63

There are 1 1 2n band members in each row. The band has 63 members.

65. a. a1 5 4, d 5 8

an 5 4 1 (n 2 1)(8)

an 5 24 1 8n

b. S12 5 12 1 a1 1 a12 }

2 2

S12 5 12 1 4 1 92 }

2 2

S12 5 576

There are 576 visible blocks.

66. a. n d(n)

1 D(1) 2 D(0) 5 16(12) 2 16(02) 5 16

2 D(2) 2 D(1) 5 16(22) 2 16(12) 5 48

3 D(3) 2 D(2) 5 16(32) 2 16(22) 5 80

4 D(4) 2 D(3) 5 16(42) 2 16(32) 5 112

b. a1 5 16, d 5 32

an 5 16(n 2 1)(32) 5 216 1 32n

So, a rule is d(n) 5 216 1 32n.

c.

12

n21

d(n)


n2ws-1200-a.indd 677 6/27/06 11:31:55 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.



67. S5 5 1000, n 5 5, d 5 50

an 5 a1 1 (n 2 1)d

a5 5 a1 1 (5 2 1)(50)

a5 5 a1 1 200

S5 5 5 1 a1 1 a5 }

2 2

1000 5 5 1 a1 1 a1 1 200 }}

2 2

1000 1 2 } 5 2 5 2a1 1 200

400 2 200 5 2a1

100 5 a1

$100 should be given away on the fi rst day.

68. a. n dn (in.) ln (in.)

1 2 2π

2 2.008 2.008π

3 2.016 2.016π

4 2.024 2.024π

Sample calculation:

d2 5 2 1 0.004(1)(2) 5 2.008

b. l1, l2, l3, . . . is an arithmetic sequence.

ln 5 l1 1 (n 2 1)(d)

ln 5 2π 1 (n 2 1)(0.008π)

ln 5 [2 1 (n 2 1)(0.008)]π

c. a1 5 2.008, d 5 0.008, an 5 5

an 5 a1 1 (n 2 1)d

5 5 2.008 1 (n 2 1)(0.008)

3 5 0.008n

375 5 n

The paper must be wrapped around the dowel 375 times.

ln 5 [2 1 (n 2 1)(0.008)]π (from part b)

l375 5 4.992π

l1 5 2π

S375 5 375 1 2π 1 4.992π }

2 2 ø 4118.6

The length of paper in the roll is about 4118.6 inches.

d. For a roll with a 7-inch diameter:

7 5 2.008 1 (n 2 1)(0.008)

5 5 0.008n

625 5 n

l1 5 2π

l625 5 [2 1 (625 2 1)(0.008)]π 5 6.992π

S625 5 625 1 2π 1 6.992π }

2 2 ø 8827.9

The length of paper in the 7-inch diameter roll is about 8827.9 inches. This is about 2.14 times the length of paper in the 5-inch diameter roll, so you could expect to pay about 2.14($1.50) 5 $3.21 for the 7-inch diameter roll.

Mixed Review

70. x1/5 5 7 71. x2/3 5 36

(x1/5)5 5 75 (x2/3)3/2 5 363/2

x 5 16,807 x 5 216

72. 6x2/5 2 5 5 19

6x2/5 5 24

x2/5 5 4

(x2/5)5/2 5 45/2

x 5 32

73. 3x3/5 2 4 5 5

3x3/5 5 9

x3/5 5 3

(x3/5)5/3 5 35/3

x 5 3 p 32/3

x 5 3 3 Ï}

32

x 5 3 3 Ï}

9

74. (x 1 10)1/4 5 5 75. (x 2 3)3/4 5 64

[(x 1 10)1/4]4 5 54 [(x 2 3)3/4]4/3 5 644/3

x 1 10 5 625 x 2 3 5 256

x 5 615 x 5 259

76. 6x 5 216 77. 5x 5 32

log6 6x 5 log6 216 log5 5

x 5 log5 32

x 5 log6 216 x 5 log5 32

x 5 log 216

} log 6

x 5 log 32

} log 5

x 5 3 x ø 2.153

78. 104x 2 6 5 12 79. 105x 1 1 1 5 5 19

104x 5 18 105x 1 1 5 14

log 104x 5 log 18 log 105x 1 1 5 log 14

4x 5 log 18 5x 1 1 5 log 14

x ø 0.314 x ø 0.029

80. 7x 2 3 5 49x 2 8

7x 2 3 5 (72)x 2 8

7x 2 3 5 72x 2 16

x 2 3 5 2x 2 16

13 5 x

81. 34x 1 1 5 729x

34x 1 1 5 (36)x

34x 1 1 5 36x

4x 1 1 5 6x

1 5 2x

1 }

2 5 x

n2ws-1200-a.indd 678 6/27/06 11:31:59 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.

679Algebra 2


82. Mean: } x 5 5 1 6 1 6 1 6 1 8 1 9

}} 6 5 40

} 6 5 6 2 }

3

Median: 6

Mode: 6

83. Mean: } x 5 32 1 36 1 38 1 43 1 43 1 45 1 46

}}} 7 5 283

} 7

ø 40.43

Median: 43

Mode: 43

84. Mean: } x 5 75 1 80 1 81 1 82 1 83 1 92 1 92 1 92

}}}} 8

5 677

} 8 5 84.625

Median: 82 1 83

} 2 5 82.5

Mode: 92

85. Mean: } x 5 25 1 (24) 1 (23) 1 (22) 1 (21) 1 1 1 5

}}}} 7

5 29

} 7 ø 21.29

Median: 22

Mode: none

86. Mean: } x 5 1.9 1 1.9 1 2.5 1 2.6 1 2.8 1 3.1 1 3.5

}}}} 7

5 18.3

} 7 ø 2.61

Median: 2.6

Mode: 1.9

87. Mean: } x 5 0.9 1 1.8 1 2.5 1 3.8 1 4.2 1 5.2 1 6.7

}}}} 7

5 25.1

} 7 ø 3.59

Median: 3.8

Mode: none

88. h 5 number of hats

s 5 number of scarves

16h 1 18s 5 710 16h 1 18s 5 710

h 1 s 5 42 216h 2 16s 5 2672

2s 5 38

s 5 19

h 1 19 5 42 → h 5 23

You sold 23 hats.

Lesson 12.3


1. a2

} a1 5

27 } 81 5

1 } 3

a3

} a2 5

9 } 27 5

1 } 3

a4

} a3 5

3 } 9 5

1 } 3

a5

} a4 5

1 } 3

The series is geometric with a common ratio of 1 }

3 .

2. a2

} a1 5

2 } 1 5 2

a3

} a2 5

6 } 2 5 3

Because there is no common ratio, the sequence is not geometric.

3. a2

} a1 5

8 }

24 5 22

a3

} a2 5

216 } 8 5 22

a4

} a3 5

32 }

216 5 22

a5

} a4 5

264 } 32 5 22

The series is geometric with a common ratio of 22.

4. a1 5 3, r 5 15

} 3 5 5

an 5 a1rn 2 1

an 5 3(5)n 2 1

a8 5 3(5)8 2 1

a8 5 234,375

5. a6 5 296, r 5 2

a6 5 a1r6 2 1

296 5 a1(2)5

23 5 a1

an 5 a1(r)n 2 1

an 5 23(2)n 2 1

a8 5 23(2)8 2 1

a8 5 2384

6. a2 5 212, a4 5 23

a2 5 a1r2 2 1 → 212 5 a1r → a1 5 2 12

} r

a4 5 a1r4 2 1 → 23 5 a1r3

23 5 1 212 } r 2 r 3

23 5 212r2

1 }

4 5 r2

6 1 } 2 5 r


n2ws-1200-a.indd 679 6/27/06 11:32:03 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.


If r 5 1 } 2 :

a1 5 212

} 1 }

2 5 224

an 5 a1rn 2 1

an 5 224 1 1 } 2 2

n 2 1

a8 5 224 1 1 } 2 2 8 2 1

5 20.1875

If r 5 2 1 } 2 :

a1 5 212

} 2

1 } 2 5 24

an 5 a1rn 2 1

an 5 24 1 2 1 } 2 2 n 2 1

a8 5 24 1 2 1 } 2 2 7 5 20.1875

7. S8 5 a1 1 1 2 r8 }

1 2 r 2

S8 5 6 1 1 2 (22)8

} 1 2 (22)

2 S8 5 2510

8. an 5 5.02(1.059)n 2 1

a11 5 5.02(1.059)11 2 1

a11 ø 8.906

The box offi ce revenue in 2000 was about $8.91 billion.

12.3 Exercises (pp. 814–817)

Skill Practice

1. The constant ratio of consecutive terms in a geometric sequence is called the common ratio.

2. If you divide each consecutive term and get the same ratio every time, the sequence is geometric.

3. a2

} a1

5 4 } 1 5 4

a3

} a2

5 8 } 4 5 2

Because there is no common ratio, the sequence is not geometric.

4. a2

} a1

5 16

} 4 5 4

a3

} a2

5 64

} 16 5 4

a4

} a3

5 256

} 64 5 4

a5

} a4

5 1024

} 256 5 4

The sequence is geometric with a common ratio of 4.

5. a2

} a1

5 36

} 216 5 1 } 6

a3

} a2

5 6 } 36 5

1 } 6

a4

} a3

5 1 } 6

a5

} a4

5 1 } 6

The sequence is geometric with a common ratio of 1 } 6 .

6. a2

} a1

5 2 } 3 }

1 }

3 5 2

a3

} a2

5 4 } 3 }

2 }

3 5 2

a4

} a3

5 8 } 3 }

4 }

3 5 2

a5

} a4

5 16

} 3 }

8 }

3 5 2


7. a2 } a1

5 1

} 1 } 2 5 2

a3 } a2

5 3 } 2 } 1 5

3 } 2

The sequence is not geometric because there is no common ratio.

8. a2 } a1

5 3 } 8 }

2 1 } 4 5 2

3 } 2

a3 } a2

5 2

3 } 16 }

3 }

8 5 2

1 } 2


9. a2 } a1

5 5 } 10 5 0.5

a3 } a2

5 2.5

} 5 5 0.5

a4 } a3

5 1.25

} 2.5 5 0.5

a5 } a4

5 0.625

} 1.25 5 0.5

The sequence is geometric with a common ratio of 0.5.

10. a2

} a1 5

26 }

23 5 2

a3

} a2 5

12 }

26 5 22



n2ws-1200-a.indd 680 6/27/06 11:32:08 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.

681Algebra 2


11. a2

} a1 5

12 }

24 5 23

a3

} a2 5

236 } 12 5 23

a4

} a3 5

108 }

236 5 23

a5

} a4 5

2324 } 108 5 23


12. a2

} a1 5

0.6 } 0.2 5 3

a3

} a2 5

1.8 } 0.6 5 3

a4

} a3 5

5.4 } 1.8 5 3

a5

} a4 5

16.2 } 5.4 5 3


13. a2

} a1 5

10 }

25 5 22

a3

} a2 5

20 } 10 5 2


14. a2

} a1 5

1.5 } 0.75 5 2

a3

} a2 5

2.25 } 1.5 5 1.5


15. a1 5 1, r 5 2 4 } 1 5 24 16. a1 5 6, r 5

18 } 6 5 3

an 5 1(24)n 2 1 an 5 6(3)n 2 1

a7 5 (24)7 2 1 5 4096 a7 5 6(3)7 2 1 5 4374

17. a1 5 4, r 5 24

} 4 5 6 18. a1 5 7, r 5 2 35

} 7 5 25

an 5 4(6)n 2 1 an 5 7(25)n 2 1

a7 5 4(6)7 2 1 a7 5 7(25)7 2 1

5 186,624 5 109,375

19. a1 5 2, r 5 3 } 2 } 2 5

3 } 4

an 5 2 1 3 } 4 2 n 2 1

a7 5 2 1 3 } 4 2 7 2 1

5 729

} 2048

20. a1 5 3, r 5 2

6 } 5 } 3 5 2

2 } 5

an 5 3 1 2 2 } 5 2 n 2 1

a7 5 3 1 2 2 } 5 2 7 2 1

5 192

} 15,625

21. a1 5 4, r 5 2 } 4 5

1 } 2

an 5 4 1 1 } 2 2 n 2 1

a7 5 4 1 1 } 2 2 7 2 1

5 1 } 16

22. a1 5 20.3, r 5 0.6

} 20.3 5 22

an 5 20.3(22)n 2 1

a7 5 20.3(22)7 2 1 5 219.2

23. a1 5 22, r 5 20.8

} 22 5 0.4

an 5 22(0.4)n 2 1

a7 5 22(0.4)7 2 1 5 20.008192

24. a1 5 7, r 5 24.2

} 7 5 20.6

an 5 7(20.6)n 2 1

a7 5 7(20.6)7 2 1 5 0.326592

25. a1 5 5, r 5 214

} 5 5 22.8

an 5 5(22.8)n 2 1

a7 5 5(22.8)7 2 1 5 2409.45152

26. a1 5 120, r 5 180

} 120 5 1.5

an 5 120(1.5)n 2 1

a7 5 120(1.5)7 2 1 5 1366.875

27. B;

a1 5 5, r 5 20

} 5 5 4

an 5 5(4)n 2 1

28. a1 5 5, r 5 3

an 5 5(3)n 2 1

n 1 2 3 4 5 6

an 5 15 45 135 405 1215

125

n21

an


n2ws-1200-a.indd 681 6/27/06 11:32:13 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.


29. a1 5 22, r 5 6

an 5 22(6)n 2 1

n 1 2 3 4 5 6

an 22 212 272 2432 22592 215,552

1600

n1

an

30. a2 5 6, r 5 2

a2 5 a1r2 2 1

6 5 a1(2)1

3 5 a1

an 5 3(2)n 2 1

n 1 2 3 4 5 6

an 3 6 12 24 48 96

10

n21

an

31. a2 5 15, r 5 1 } 2

a2 5 a1r2 2 1

15 5 a1 1 1 } 2 2 1

30 5 a1

an 5 30 1 1 } 2 2 n 2 1

n 1 2 3 4 5 6

an 30 15 7.5 3.75 1.875 0.9375

3

n1

an

32. a5 5 1, r 5 1 } 8

a5 5 a1r5 2 1

1 5 a1 1 1 } 8 2 4

4096 5 a1

an 5 4096 1 1 } 8 2

n 2 1

n 1 2 3 4 5 6

an 4096 512 64 8 1 0.125

400

n21

an

33. a4 5 212, r 5 2 1 } 4

a4 5 a1r4 2 1

212 5 a1 1 2 1 } 4 2 3

768 5 a1

an 5 768 1 2 1 } 4 2 n 2 1

n 1 2 3 4 5 6

an 768 2192 48 212 3 20.75

100

n21

an


n2ws-1200-a.indd 682 6/27/06 11:32:23 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.

683Algebra 2


34. a3 5 75, r 5 5

a3 5 a1r3 2 1

75 5 a1(5)2

3 5 a1

an 5 3(5)n 2 1

n 1 2 3 4 5 6

an 3 15 75 375 1875 9375

1000

n21

an

35. a2 5 8, r 5 4

a2 5 a1r2 2 1

8 5 a1(4)1

2 5 a1

an 5 2(4)n 2 1

n 1 2 3 4 5 6

an 2 8 32 128 512 2048

200

n21

an

36. a4 5 500, r 5 5

a4 5 a1r4 2 1

500 5 a1(5)3

4 5 a1

an 5 4(5)n 2 1

n 1 2 3 4 5 6

an 4 20 100 500 2500 12,500

1500

n21

an

37. In the fi rst step, the exponent should be n 2 1. an 5 a1rn 2 1

an 5 3(2)n 2 1

38. In the fi rst step, r should be raised to the (n 2 1)th power, not a1.

an 5 a1rn 2 1

an 5 3(2)n 2 1

39. a1 5 3, a3 5 12

a3 5 a1r3 2 1

12 5 3r2

4 5 r2

62 5 r

an 5 3(2)n 2 1 or an 5 3(22)n 2 1

40. a1 5 1, a5 5 625

a5 5 a1r5 2 1

625 5 1 p r4

65 5 r

an 5 1(5)n 2 1 or an 5 1(25)n 2 1

41. a1 5 2 1 } 4 , a4 5 216

a4 5 a1r4 2 1

216 5 2 1 } 4 r3

64 5 r3

4 5 r

an 5 2 1 } 4 (4)n 2 1

42. a3 5 10, a6 5 270

a3 5 a1r3 2 1 → 10 5 a1r2 → a1 5 10

} r2

a6 5 a1r6 2 1 → 270 5 a1r5

270 5 1 10 }

r2 2 r5

270 5 10r3

27 5 r3

3 5 r

a1 5 10

} 32 5 10

} 9

an 5 a1r n 2 1

an 5 10

} 9 (3)n 2 1


n2ws-1200-a.indd 683 6/27/06 11:32:29 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.


43. a2 5 240, a4 5 210

a2 5 a1r2 2 1 → 240 5 a1r → a1 5 240

} r

a4 5 a1r4 2 1 → 210 5 a1r3

210 5 1 240 } r 2 r3

210 5 240r2

1 }

4 5 r2

6 1 } 2 5 r

a1 5 240

} 1 }

2 5 280 or a1 5

240 }

2 1 } 2 5 80

an 5 a1r n 2 1 an 5 a1rn 2 1

an 5 280 1 1 } 2 2 n 2 1 an 5 80 1 2

1 }

2 2 n 2 1

44. a2 5 224, a5 5 1536

a2 5 a1r2 2 1 → 224 5 a1r → a1 5 224

} r

a5 5 a1r5 2 1 → 1536 5 a1r4

1536 5 1 224 } r 2 r4

1536 5 224r3

264 5 r3

24 5 r

a1 5 224

} 24 5 6

an 5 a1r n 2 1

an 5 6(24)n 2 1

45. a4 5 162, a7 5 4374

a4 5 a1r4 2 1 → 162 5 a1r3 → a1 5 162

} r3

a7 5 a1r7 2 1 → 4374 5 a1r6

4374 5 1 162 }

r3 2 r6

4374 5 162r3

27 5 r2

3 5 r

a1 5 162

} 33 5 6

an 5 a1r n 2 1

an 5 6(3)n 2 1

46. a3 5 7 } 4 , a5 5

7 } 16

a3 5 a1r3 2 1 → 7 }

4 5 a1r2 → a1 5

7 }

4r2

a5 5 a1r5 2 1 → 7 }

16 5 a1r4

7 }

16 5 1 7

} 4r2 2 r4

7 }

16 5

7 } 4 r2

1 }

4 5 r2

6 1 } 2 5 r

a1 5 7 } 4 }

1 6 1 } 2 2 2

5 7

an 5 a1r n 2 1

an 5 7 1 1 } 2 2

n 2 1 or an 5 7 1 2

1 } 2 2 n 2 1

47. a4 5 6, a7 5 243

} 8

a4 5 a1r4 2 1 → 6 5 a1r3 → a1 5 6 }

r3

a7 5 a1r7 2 1 → 243

} 8 5 a1r6

243

} 8 5 1 6 }

r3 2 r6

243

} 8 5 6r3

243

} 48

5 r3

3 Î}

243

} 48

5 r

3

3 Ï}

12 }

4 5 r

a1 5 6 }

1 3 3 Ï}

12 } 4 2 3

5 32

} 27

an 5 a1r n 2 1

an 5 32

} 27 1 3 3 Ï}

12 }

4 2 n 2 1

48. S10 5 a1 1 1 2 r10 }

1 2 r 2

S10 5 5 1 1 2 210 }

1 2 2 2 5 5115

49. S8 5 a1 1 1 2 r8 }

1 2 r 2

S8 5 6 1 1 2 48 }

1 2 4 2 5 131,070


n2ws-1200-a.indd 684 6/27/06 11:32:34 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.

685Algebra 2


50. i 5 0

∑ 7

12 1 2 1 } 2 2 i 5

i 5 1 ∑

8

12 1 2 1 } 2 2 i 2 1

S8 5 a1 1 1 2 r8 }

1 2 r 2

S8 5 12 1 1 2 1 2 1 } 2 2 8 }

1 2 1 2 1 } 2 2 2 5

255 } 32

51. S6 5 a1 1 1 2 r6 }

1 2 r 2

S6 5 4 1 1 2 1 1 } 4 2 6 }

1 2 1 1 } 4 2 2 5

1365 } 256

52. S12 5 a1 1 1 2 r12 }

1 2 r 2

S12 5 8 1 1 2 1 3 } 2 2 12

} 1 2 1 3 } 2 2

2 5 527,345

} 256

53. i 5 0

∑ 10

(24)i 5 i 5 1

∑ 11

(24)i 2 1

S11 5 a1 1 1 2 r11 }

1 2 r 2

S11 5 1 1 1 2 (24)11

} 1 2 (24)

2 5 838,861

54. C;

S9 5 a1 1 1 2 r9 }

1 2 r 2

S9 5 2 1 1 2 39 }

1 2 3 2 5 19,682

55. Sample answer:

S5 5 100, choose r 5 2, fi nd a1.

S5 5 a1 1 1 2 r5 }

1 2 r 2

100 5 a1 1 1 2 25 }

1 2 2 2

100 5 a1(31)

100

} 31

5 a1

Series: 100

} 31

1 200

} 31 1 400

} 31 1 800

} 31 1 1600

} 31

56. a. a1 5 1, r 5 x } 1 5 x

S5 5 a1 1 1 2 r5 }

1 2 r 2

S5 5 1 1 1 2 x5 }

1 2 x 2 5

1 2 x5

} 1 2 x

b. a1 5 3x, r 5 6x3

} 3x 5 2x2

S4 5 a1 1 1 2 r4 }

1 2 r 2

S4 5 3x 1 1 2 (2x2)4

} 1 2 2x2 2

S4 5 3x 1 1 2 16x8 }

1 2 2x2 2

Problem Solving

57. a. a1 5 5, r 5 2

an 5 5(2)n 2 1

b. S4 5 a1 1 1 2 r4 }

1 2 r 2

S4 5 5 1 1 2 24 }

1 2 2 2 5 75

There are 75 skydivers in four rings.

58. a. a1 5 32, r 5 1 } 2

an 5 32 1 1 } 2 2 n 2 1

1 ≤ n ≤ 6 are the logical values because a6 5 1.

b. S6 5 a1 1 1 2 r6 }

1 2 r 2

S6 5 32 1 1 2 1 1 } 2 2 6 }

1 2 1 } 2 2 5 63

63 games are played in the tournament.

59. a. After the fi rst pass, 512 items remain, so a1 5 512

and r 5 1 } 2 .

an 5 512 1 1 } 2 2 n 2 1

b. Find n when an 5 1.

an 5 512 1 1 } 2 2 n 2 1

1 5 512 1 1 } 2 2 n 2 1

1 }

512 5 1 1 } 2 2 n 2 1

log 1 }

512 5 (n 2 1)log 1 1 }

2 2

log

1 }

512 }

log 1 }

2 1 1 5 n

10 5 n

On the tenth pass, only one term remains.

60. a. a1 5 1, r 5 8

an 5 1(8)n 2 1

an 5 8n 2 1

S8 5 a1 1 1 2 r8 }

1 2 r 2

S8 5 1 1 1 2 88 }

1 2 8 2 5 2,396,745

2,396,745 squares are removed through stage 8.

b. b1 5 1 2 1 } 9 5

8 } 9 , r 5

8 } 9

bn 5 8 } 9 1 8 }

9 2 n 2 1

b12 5 8 } 9 1 8 }

9 2 12 2 1

ø 0.2433

The remaining area of the original square after the twelfth stage is about 0.2433 square unit.


n2ws-1200-a.indd 685 6/27/06 11:32:39 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.


61. a. For Company A:

a1 5 20,000, d 5 1000

an 5 a1 1 (n 2 1)d

an 5 20,000 1 (n 2 1)(1000)

an 5 19,000 1 1000n

This sequence is arithmetic.

For Company B:

b1 5 20,000, r 5 1.04

bn 5 a1r n 2 1

bn 5 20,000(1.04)n 2 1

This sequence is geometric.

b.

Years of employment

Sal

ary

(do

llars

)

bn

an

19,0000

20,00021,00022,00023,00024,00025,00026,00027,000

n1 2 3 4 5 6 70

c. For Company A:

Sn 5 n 1 a1 1 an }

2 2

S20 5 20 1 20,000 1 39,000 }}

2 2 5 590,000

For Company B:

Sn 5 a1 1 1 2 rn }

1 2 r 2

S20 5 20,000 1 1 2 (1.04)20

} 1 2 1.04

2 5 595,561.57

The sum of the wages during the fi rst 20 years of employment is $590,000 for Company A and about $595,562 for Company B.

d.

For Company A

n an Sn

1 $20,000.00 $20,000.00

2 $21,000.00 $41,000.003 $22,000.00 $63,000.004 $23,000.00 $86,000.005 $24,000.00 $110,000.006 $25,000.00 $135,000.007 $26,000.00 $161,000.008 $27,000.00 $188,000.009 $28,000.00 $216,000.0010 $29,000.00 $245,000.0011 $30,000.00 $275,000.0012 $31,000.00 $306,000.0013 $32,000.00 $338,000.0014 $33,000.00 $371,000.0015 $34,000.00 $405,000.0016 $35,000.00 $440,000.0017 $36,000.00 $476,000.0018 $37,000.00 $513,000.0019 $38,000.00 $551,000.0020 $39,000.00 $590,000.00

For Company B

n Sn

1 $20,000.00

2 $40,800.003 $62,432.004 $84,929.285 $108,326.456 $132,659.517 $157,965.898 $184,284.539 $211,655.9110 $240,122.1411 $269,727.0312 $300,516.1113 $332,536.7514 $365,838.2215 $400,471.7516 $436,490.6217 $473,950.2518 $512,908.2619 $553,424.5920 $595,561.57

The total amount earned by Company B is greater than the amount earned by Company A after 19 years.

62. a1 5 2000, r 5 1.05

S30 5 a1 1 1 2 r30 }

1 2 r 2

S30 5 2000 1 1 2 1.0530 }

1 2 1.05 2 ø 132,877.70

You will have $132,877.70 in the IRA after your last deposit.

Mixed Review

63.

10 221

13

29

12

56

64.

2 3 4 51023 22 21

472 6

65.

2 31023 22 212627 25 24

1142 8 2.721.8

66. 4 }

1 1 x 5 9 67.

3 }

1 2 x 5 10

4 5 9(1 1 x) 3 5 10(1 2 x)

4 5 9 1 9x 3 5 10 2 10x

25 5 9x 27 5 210x

2 5 } 9 5 x x 5

7 } 10


n2ws-1200-a.indd 686 6/27/06 11:32:46 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.

687Algebra 2


68. 2 }

3x 5

5 } x 1 4

2(x 1 4) 5 5(3x)

2x 1 8 5 15x

8 5 13x

8 }

13 5 x

69. x }

x 2 6 5

x } 3

x(3) 5 x(x 2 6)

3x 5 x2 2 6x

0 5 x2 2 9x

0 5 x(x 2 9)

x 5 0 or x 5 9

70. 2 18

} x 2 x 5 11

x 1 2 18

} x 2 x 2 5 x(11)

218 2 x2 5 11x

0 5 x2 1 11x 1 18

0 5 (x 1 2)(x 1 9)

x 5 22 or x 5 29

71. x 1 16 5 x2

} x 2 8

(x 2 8)(x 1 16) 5 x2

x2 1 16x 2 8x 2 128 5 x2

8x 2 128 5 0

8x 5 128

x 5 16

72. i 5 2

∑ 8

(i2 1 1) 5 (22 1 1) 1 (32 1 1) 1 (42 1 1) 1 (52 1 1) 1 (62 1 1) 1 (72 1 1) 1 (82 1 1) 5 5 1 10 1 17 1 26 1 37 1 50 1 65

5 210

73. i 5 3

∑ 11

6i 5 6(3) 1 6(4) 1 6(5) 1 6(6) 1 6(7)

1 6(8) 1 6(9) 1 6(10) 1 6(11)

5 18 1 24 1 30 1 36 1 42 1 48 1 54 1 60 1 66

5 378

74. i 5 1

∑ 6

(5 2 i) 5 (5 2 1) 1 (5 2 2) 1 (5 2 3)

1 (5 2 4) 1 (5 2 5) 1 (5 2 6)

5 4 1 3 1 2 1 1 1 0 1 (21)

5 9

75. i 5 7

∑ 15

(3i 1 4) 5 (3(7) 1 4) 1 (3(8) 1 4)

1 (3(9) 1 4) 1 . . . 1 (3(14) 1 4)

1 (3(15) 1 4)

5 25 1 28 1 31 1 34 1 37 1 40

1 43 1 46 1 49

5 333

76. i 5 5

∑ 12

(22i 1 1) 5 (22(5) 1 1) 1 (22(6) 1 1) 1 . . .

1 (22(11) 1 1) 1 (22(12) 1 1)

5 29 2 11 2 13 2 15 2 17

2 19 2 21 2 23

5 2128

77. i 5 4

∑ 9

4i2 5 4(4)2 1 4(5)2 1 4(6)2 1 4(7)2 1 4(8)2 1 4(9)2

5 64 1 100 1 144 1 196 1 256 1 324

5 1084

Quiz 12.1–12.3 (p. 817)

1. Given terms: 1, 3, 5, 7, . . .

Rewritten terms: 2(1) 2 1, 2(2) 2 1, 2(3) 2 1, 2(4) 2 1

Next term: 2(5) 2 1 5 9

Rule for nth term: an 5 2n 2 1

2. Given terms: 25, 10, 215, 20, . . .

Rewritten terms: (21)1(5(1)), (21)2(5(2)), (21)3(5(3)), (21)4(5(4))

Next term: (21)5(5(5)) 5 225

Rule for nth term: an 5 (21)n(5n)

3. Given terms: 1 }

20 , 2 }

30 , 3 }

40 , 4 }

50 , . . .

Rewritten terms: 1 }

(1 1 1)(10) ,

2 }

(2 1 1)(10) ,

3 }

(3 1 1)(10) ,

4 }

(4 1 1)(10)

Next term: 5 }

(5 1 1)(10) 5

5 } 60

Rule for nth term: an 5 n }

10(n 1 1)

4. Given terms: 4, 16, 64, 256, . . .

Rewritten terms: 41, 42, 43, 44

Next term: 45 5 1024

Rule for nth term: an 5 4n

5. Given terms: 2, 6, 12, 20, . . .

Rewritten terms: 1(1 1 1), 2(2 1 1), 3(3 1 1), 4(4 1 1)

Next term: 5(5 1 1) 5 30

Rule for nth term: an 5 n(n 1 1)

6. Given terms: 9, 36, 81, 144, . . .

Rewritten terms: 9(12), 9(22), 9(32), 9(42) Next term: 9(52) 5 225

Rule for nth term: an 5 9n2 5 (3n)2

7. i 5 1

∑ 4

2i3 5 2(13) 1 2(23) 1 2(33) 1 2(43) 5 2 1 16 1 54 1 128

5 200

8. k 5 1

∑ 5

(k2 1 3) 5 (12 1 3) 1 (22 1 3) 1 (32 1 3) 1 (42 1 3) 1 (52 1 3) 5 4 1 7 1 12 1 19 1 28

5 70


n2ws-1200-a.indd 687 6/27/06 11:32:50 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.


9. n 5 2

∑ 6

1 }

n 2 1 5

1 } 2 2 1 1

1 } 3 2 1 1

1 } 4 2 1 1

1 } 5 2 1 1

1 } 6 2 1

5 1 1 1 } 2 1

1 } 3 1

1 } 4 1

1 } 5

5 137

} 60

10. a1 5 1, d 5 6

an 5 a1 1 (n 2 1)d

an 5 1 1 (n 2 1)(6)

an 5 25 1 6n

a15 5 25 1 6(15) 5 85

Sn 5 n 1 a1 1 a15 } 2 2

S15 5 15 1 1 1 85 }

2 2 5 645

11. a1 5 1 } 2 , d 5

3 } 2

an 5 a1 1 (n 2 1)d

an 5 1 } 2 1 (n 2 1) 1 3 }

2 2

an 5 21 1 3 } 2 n

a15 5 21 1 3 } 2 (15) 5

43 } 2

Sn 5 n 1 a1 1 an } 2 2

S15 5 15 1 1 }

2 1

43 } 2 }

2 2 5 165

12. a1 5 5, d 5 23

an 5 a1 1 (n 2 1)d

an 5 5 1 (n 2 1)(23)

an 5 8 2 3n

a15 5 8 2 3(15) 5 237

Sn 5 n 1 a1 1 an } 2 2

S15 5 15 1 5 1 (237) }

2 2 5 2240

13. a1 5 2, r 5 8 } 2 5 4

an 5 a1rn 2 1

an 5 2(4)n 2 1

a15 5 2(4)15 2 1 5 536,870,912

Sn 5 a1 1 1 2 rn }

1 2 r 2

S15 5 2 1 1 2 415 }

1 2 4 2 5 715,827,882

14. a1 5 2, r 5 4 } 3 } 2 5

2 } 3

an 5 a1rn 2 1

an 5 2 1 2 } 3 2

n 2 1

a15 5 2 1 2 } 3 2

15 2 1 5

32,768 } 4,782,969

Sn 5 a1 1 1 2 rn }

1 2 r 2

S15 5 2 1 1 2 1 2 } 3 2 15

} 1 2 1 2 } 3 2

2 ø 5.986

15. a1 5 23, r 5 15

} 23 5 25

an 5 a1rn 2 1

an 5 23(25)n 2 1

a15 5 23(25)15 2 1 5 218,310,546,875

Sn 5 a1 1 1 2 rn }

1 2 r 2

S15 5 23 1 1 2 (25)15

} 1 2 (25)

2 5 215,258,789,063

16. a1 5 2057, r 5 1.06

an 5 a1rn 2 1

an 5 2057(1.06)n 2 1

In 2002, n 5 8.

a8 5 2057(1.06)8 2 1 ø 3092.97

In 2002, the average tuition at a public college was about $3092.97.

Mixed Review of Problem Solving (p. 818)

1. a. a1 5 45,000, r 5 1.035

an 5 a1rn 2 1

an 5 45,000(1.035)n 2 1

b. a5 5 45,000(1.035)5 2 1 ø 51,638.54

During your 5th year of employment, your salary will be $51,638.54.

c. Sn 5 a1 1 1 2 rn }

1 2 r 2

S30 5 45,000 1 1 2 (1.035)30

} 1 2 1.035

2 ø 2,323,020.48

After 30 years, you will have earned a total of $2,323,020.48.

2. a. A 5 πr2

a1 5 π(12) 5 π

a2 5 π(22) 2 π(12) 5 3π

a3 5 π(32) 2 π(22) 5 5π

a1 5 π, d 5 2π

an 5 π 1 (n 2 1)(2π) 5 (2n 2 1)π

b. i 5 1

∑ n

(2i 2 1)π


n2ws-1200-a.indd 688 6/27/06 11:32:55 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.

689Algebra 2


c. For n 5 1: i 5 1

∑ 1

(2i 2 1)π 5 π

For n 5 2: i 5 1

∑ 2

(2i 2 1)π 5 π 1 3π 5 4π

For n 5 4: i 5 1

∑ 4

(2i 2 1)π 5 π 1 3π 1 5π 1 7π

5 16π

For n 5 8:

i 5 1

∑ 8

(2i 2 1)π 5 π 1 3π 1 5π 1 7π 1 9π

1 11π 1 13π 1 15π 5 64π

When the number of rings is doubled, the total area is quadrupled.

3. Tables placed together by short edges:

Sequence: 6, 10, 14, 18 . . .

a1 5 6, d 5 4

an 5 6 1 (n 2 1)(4) 5 2 1 4n

Tables placed together by long edges:

Sequence: 6, 8, 10, 12 . . .

a1 5 6, d 5 2

an 5 6 1 (n 2 1)(2) 5 4 1 2n

When the tables are connected by the short edges, 2 1 4n 2 (4 1 2n), or 22 1 2n, more people can be seated.

4. Sample answer:

n 5 8, Sn 5 70

Sn 5 n 1 a1 1 an } 2 2

70 5 8 1 a1 1 a8 } 2 2

17.5 5 a1 1 a8

Choose a1 5 1.75 and a8 5 15.75.

an 5 a1 1 (n 2 1)d

a8 5 a1 1 (8 2 1)d

15.75 5 1.75 1 7d

14 5 7d

2 5 d

A series is i 5 1

∑ 8

(2i 2 0.25). So, an 5 1.75 1 (n 2 1)(2),

or an 5 20.25 1 2n.

5. a1 5 15, d 5 21

an 5 a1 1 (n 2 1)d

an 5 15 1 (n 2 1)(21)

an 5 16 2 n

6 5 16 2 n

n 5 10

Sn 5 n 1 a1 1 an }

2 2

S10 5 10 1 15 1 6 }

2 2 5 105

There are 105 pieces of chalk in the pile.

6. a1 5 9, d 5 7

an 5 a1 1 (n 2 1)d

an 5 9 1 (n 2 1)(7)

an 5 2 1 7n

The height at the top of the 10th stair is 2 1 7(10) 5 72 inches. To fi nd the height of the bottom of the nth stair, subtract the height of the nth stair from an. So, an 5 2 1 7n 2 7 5 25 1 7n.

7. a. The sequence is geometric because each term is half of the previous term.

b. a1 5 66, r 5 1 } 2

an 5 a1rn 2 1

an 5 66 1 1 } 2 2 n 2 1

c.

n 1 2 3 4 5 6 7

an 66 33 16.5 8.25 4.125 2.0625 1.03125

7

n

an

1

The points lie on an exponential decay curve.

d. an < 1

66 1 1 } 2 2 n 2 1

< 1

1 1 } 2 2 n 2 1

< 1 }

66

(n 2 1) log 1 1 } 2 2 < log

1 }

66

n 2 1 < log

1 }

66 }

log 1 }

2

n < 7.044

Because n represents the number of two-hour intervals, there will be less than one gram of Platinum-197 after about 14 hours.


n2ws-1200-a.indd 689 6/27/06 11:32:59 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.


8. Sample answer:

Choose a geometric series with fi ve terms: 1 1 2 1 4 1 8 1 16 5 31.

Next, fi nd an arithmetic series with fi ve terms and a sum of 31.

Sn 5 1 a1 1 an } 2 2

31 5 5 1 a1 1 a5 }

2 2

12.4 5 a1 1 a5

Choose a1 5 2.2 and a5 5 10.2.

an 5 a1 1 (n 2 1)d

10.2 5 2.2 1 4d

8 5 4d

2 5 d

So, an 5 2.2 1 (n 2 1)(2) 5 0.2 1 2n and the series is 2.2 1 4.2 1 6.2 1 8.2 1 10.2 5 31.

Lesson 12.4

Investigating Algebra Activity 12.4 (p. 819)

Step 4. The next three areas are 1 }

8 ,

1 }

16 , and

1 }

32 . These areas

form a geometric sequence because there is a common

ratio of 1 }

2 .

Step 5.

Number of pieces

1 2 3

Combined area

1 } 2

1 }

2 1

1 } 4 5

3 } 4

3 }

4 1

1 } 8 5

7 } 8

Number of pieces

4 5

Combined area

7 } 8 1

1 } 16 5

15 } 16

15 }

16 1

1 } 32 5

31 } 32

1. The total area appears to be approaching one square unit.

2. An 5 A1 1 1 2 rn }

1 2 r 2

An 5 1 } 2 1 1 2 1 1 } 2 2 n

} 1 2

1 } 2 2 5 1 2 1 1 } 2 2 n

As n gets large, An gets close to 1. For example, when n 5 10, An 5 0.9990234375, and when n 5 100, An ø 1.


1. S1 5 2 } 5 5 0.40

S2 5 2 } 5 1

4 } 25 5 0.56

S3 5 2 } 5 1

4 } 25 1

8 } 125 ø 0.62

S4 5 2 } 5 1

4 } 25 1

8 } 125 1

16 } 625 ø 0.65

S5 5 2 } 5 1

4 } 25 1

8 } 125 1

16 } 625 1

32 } 3125 ø 0.66

As n increases, Sn appears to approach 2 }

3 .

n1

0.1

Sn

2. a1 5 1, r 5 2 1 } 2

S 5 a1 } 1 2 r 5

1 }

1 2 1 2 1 } 2 2

5 2 } 3

3. r 5 5 } 4

Because 5 }

4 ≥ 1, the series has no sum.

4. a1 5 3, r 5 3 } 4 } 3 5

1 } 4

S 5 a1 } 1 2 r 5

3 }

1 2 1 } 4 5 4

5. a1 5 10, r 5 0.8

d 5 a1 } 1 2 r 5

10 } 1 2 0.8 5 50

The pendulum swings a total distance of 50 inches.

6. 0.555. . . 5 5(0.1) 1 5(0.1)2 1 5(0.1)3 1 . . .

5 a1 } 1 2 r

5 5(0.1)

} 1 2 0.1

5 0.5

} 0.9

5 5 } 9

7. 0.727272. . . 5 72(0.01) 1 72(0.01)2 1 72(0.01)3 1 . . .

5 a1 } 1 2 r

5 72(0.01)

} 1 2 0.01

5 0.72

} 0.99

5 8 } 11


n2ws-1200-a.indd 690 6/27/06 11:33:03 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.

691Algebra 2


8. 0.131313. . . 5 13(0.01) 1 13(0.01)2 1 13(0.01)3 1 . . .

5 a1 } 1 2 r

5 13(0.01)

} 1 2 0.01

5 0.13

} 0.99

5 13

} 99

12.4 Exercises (pp. 823–825)

Skill Practice

1. The sum Sn of the fi rst n terms of an infi nite series is called a partial sum.

2. If r < 1, the series has a sum.

3. S1 5 1 } 2 5 0.5

S2 5 1 } 2 1

1 } 6 ø 0.67

S3 5 1 } 2 1

1 } 6 1

1 } 18 ø 0.72

S4 5 1 } 2 1

1 } 6 1

1 } 18 1

1 } 54 ø 0.74

S5 5 1 } 2 1

1 } 6 1

1 } 18 1

1 } 54 1

1 } 162 ø 0.75

Sn appears to be approaching 3 }

4 .

0.1

x

y

1

4. S1 5 2 } 3 ø 0.67

S2 5 2 } 3 1

1 } 3 5 1

S3 5 2 } 3 1

1 } 3 1

1 } 6 ø 1.17

S4 5 2 } 3 1

1 } 3 1

1 } 6 1

1 } 12 5 1.25

S5 5 2 } 3 1

1 } 3 1

1 } 6 1

1 } 12 1

1 } 24 ø 1.29

Sn appears to be approaching 1 1 }

3 .

0.15

x

y

1

5. S1 5 4

S2 5 4 1 12

} 5 5 6.4

S3 5 4 1 12

} 5 1 36

} 25 5 7.84

S4 5 4 1 12

} 5 1 36

} 25 1 108

} 125 ø 8.70

S5 5 4 1 12

} 5 1 36

} 25 1 108

} 125 1 324

} 625 ø 9.22

Sn appears to be approaching 10.

1

x

y

1

6. S1 5 1 } 4 5 0.25

S2 5 1 } 4 1

5 } 4 5 1.5

S3 5 1 } 4 1

5 } 4 1

25 } 4 5 7.75

S4 5 1 } 4 1

5 } 4 1

25 } 4 1

125 } 4 5 39

S5 5 1 } 4 1

5 } 4 1

25 } 4 1

125 } 4 1

625 } 4 5 195.25

As n increases, Sn also increases. The series has no sum.

20

x

y

1

7. a1 5 8, r 5 1 } 5

S 5 a1 } 1 2 r 5

8 }

1 2 1 } 5 5 10

8. r 5 3 } 2

Because 3 } 2 ≥ 1, the series has no sum.

9. r 5 5 } 3

Because 5 } 3 ≥ 1, the series has no sum.

10. a1 5 11

} 3 , r 5 3 } 8

S 5 a1 } 1 2 r 5

11

} 3 }

1 2 3 } 8 5

88 } 15


n2ws-1200-b.indd 691 6/27/06 11:35:38 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.


11. a1 5 2, r 5 1 } 6

S 5 a1 } 1 2 r 5

2 }

1 2 1 } 6 5

12 } 5

12. a1 5 25, r 5 2 } 5

S 5 a1 } 1 2 r 5

25 }

1 2 2 } 5 5 2

25 } 3

13. a1 5 7, r 5 2 8 } 9

S 5 a1 } 1 2 r 5

7 }

1 2 1 2 8 } 9 2 5

63 } 17

14. r 5 2 10

} 3

Because 2 10

} 3 ≥ 1, the series has no sum.

15. r 5 4

Because 4 ≥ 1, the series has no sum.

16. a1 5 22, r 5 2 1 } 4

S 5 a1 } 1 2 r 5

22 }

1 2 1 2 1 } 4 2 5 2

8 } 5

17. a1 5 1, r 5 2 3 } 7

S 5 a1 } 1 2 r 5

1 }

1 2 1 2 3 } 7 2 5

7 } 10

18. r 5 3

Because 3 ≥ 1, the series has no sum.

19. For this series, r 5 7 } 2 .

Because 7 }


20. a1 5 2 1 } 8 , r 5

2 1 } 12 }

2 1 } 8 5

2 } 3

S 5 a1 } 1 2 r 5

2 1 } 8 }

1 2 2 } 3 5 2

3 } 8

21. a1 5 2 } 3 , r 5

2 2 } 9 }

2 } 3 5 2

1 } 3

S 5 a1 } 1 2 r 5

2 } 3 }

1 2 1 2 1 } 3 2 5

1 } 2

22. a1 5 4 } 15 , r 5

4 } 9 }

4 }

15 5

5 } 3

Because 5 }


23. a1 5 3, r 5 5 } 2 } 3 5

5 } 6

S 5 a1 } 1 2 r 5

3 }

1 2 5 } 6 5 18

24. 0.222. . . 5 2(0.1) 1 2(0.1)2 1 2(0.1)3 1 . . .

5 a1 } 1 2 r

5 2(0.1)

} 1 2 0.1

5 0.2

} 0.9

5 2 } 9

25. 0.444. . . 5 4(0.1) 1 4(0.1)2 1 4(0.1)3 1 . . .

5 a1 } 1 2 r

5 4(0.1)

} 1 2 0.1

5 0.4

} 0.9

5 4 } 9

26. 0.161616. . . 5 16(0.01) 1 16(0.01)2 1 16(0.01)3 1 . . .

5 a1 } 1 2 r

5 16(0.01)

} 1 2 0.01

5 0.16

} 0.99

5 16

} 99

27. 0.625625625. . . 5 625(0.001) 1 625(0.001)2

1 625(0.001)3 1 . . .

5 a1 } 1 2 r

5 625(0.001)

} 1 2 0.001

5 0.625

} 0.999

5 625

} 999

28. 32.3232. . . 5 32 1 32(0.01) 1 32(0.01)2 1 . . .

5 a1 } 1 2 r

5 32 } 1 2 0.01

5 32

} 0.99

5 3200

} 99

29. 130.130130. . . 5 130 1 130(0.001)

1 130(0.001)2 1 . . .

5 a1 } 1 2 r

5 130 } 1 2 0.001

5 130

} 0.999

5 130,000

} 999


n2ws-1200-b.indd 692 6/28/06 1:54:02 PM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.

693Algebra 2



30. 0.090909. . . 5 9(0.01) 1 9(0.01)2 1 9(0.01)3 1 . . .

5 a1 } 1 2 r

5 9(0.01)

} 1 2 0.01

5 0.09

} 0.99

5 1 } 11

31. 0.2777. . . 5 0.2 1 7(0.1)2 1 7(0.1)3 1 7(0.1)4 1 . . .

5 0.2 1 a1 } 1 2 r

5 0.2 1 7(0.1)2

} 1 2 0.1

5 0.2 1 0.07

} 0.9

5 18

} 90 1 7 } 90 5

5 } 18

32. C;

18.1818. . . 5 18 1 18(0.01) 1 18(0.01)2 1 . . .

5 18 } 1 2 0.01

5 1800

} 0.99

5 1800

} 99

5 200

} 11

33. 0.999. . . 5 9(0.1) 1 9(0.1)2 1 9(0.1)3 1 . . .

5 a1 } 1 2 r

5 9(0.1)

} 1 2 0.1

5 0.9

} 0.9

5 1

34. Sample answer:

a. Choose r 5 1 } 2 .

S 5 a1 } 1 2 r

5 5 a1 }

1 2 1 } 2

5 }

2 5 a1

The fi rst series is i 5 1

∑ `

5 }

2 1 1 }

2 2 i 2 1

.

b. Choose r 5 1 } 5 .

S 5 a1 } 1 2 r

5 5 a1 }

1 2 1 } 5

4 5 a1

The fi rst series is i 5 1

∑ `

4 1 1 } 5 2 i 2 1

.

35. a1 5 1, r 5 4x

} 1 5 4x

If the series has a sum, 2 1 } 4 < x <

1 }

4 .

S 5 a1 } 1 2 r 5

1 } 1 2 4x

36. a1 5 6, r 5 3 } 2 x } 6 5

1 } 4 x

If the series has a sum, 24 < x < 4.

S 5 a1 } 1 2 r 5

6 }

1 2 1 } 4 x

Problem Solving

37. d 5 14 1 14(0.8) 1 14(0.8)2 1 . . .

5 a1 } 1 2 r

5 14 } 1 2 0.8

5 70

The person swings a total distance of 70 feet.

38. S 5 350,000 1 350,000(0.88) 1 350,000(0.88)2 1 . . .

5 a1 } 1 2 r

5 350,000

} 1 2 0.88

ø 2,916,666.67

The maximum amount of profi t the company can make is $2,916,666.67.

39. D;

S 5 345 1 345(0.783) 1 345(0.783)2 1 . . .

5 a1 } 1 2 r

5 345 } 1 2 0.783

ø 1589.86

The company will ship a total of about 1.59 billion cassettes.

40. Distance: a1 5 20, r 5 10

} 20 5 1 } 2

Sd 5 a1 } 1 2 r

Sd 5 20 }

1 2 1 } 2 5 40

Time: a1 5 1, r 5 0.5

} 1 5 0.5

St 5 a1 } 1 2 r

St 5 1 } 1 2 0.5 5 2

Because both series have fi nite sums, Archilles catches up to the tortoise in 2 seconds after 40 feet is traveled.

n2ws-1200-b.indd 693 6/28/06 1:54:34 PM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.


41. a. The ball bounces 6 1 6, or 12 feet between the fi rst and second bounce, and 4.5 1 4.5, or 9 feet between the second and third bounce.

b. a1 5 12, r 5 0.75

an 5 a1rn 2 1

an 5 12(0.75)n 2 1

Series: i 5 1

∑ `

12(0.75)i 2 1

c. S 5 a1 } 1 2 r 5

12 } 1 2 0.75 5 48

The ball travels a total distance of 48 1 8 5 56 feet.

d. a1 5 2(0.75)h, r 5 0.75

Total distance 5 distance from each bounce

1 original distance

5 a1 } 1 2 r 1 h

5 2(0.75)h

} 1 2 0.75 1 h

5 6h 1 h

5 7h

If the ball is dropped from a distance of h feet, it travels a total distance of 7h feet.

42. a. a1 5 1 } 4 , a2 5

3 } 16 , r 5

3 } 16 }

1 }

4 5

3 } 4

an 5 a1rn 2 1

an 5 1 } 4 1 3 } 4 2

n 2 1

b. n 5 1

∑ `

an 5 n 5 1

∑ `

1 }

4 1 3 }

4 2

n 2 1

5 1 } 4 }

1 2 3 } 4

5 1

This answer means that eventually, one square unit of area will be removed from the triangle, so no area will remain.

Mixed Review

43. P(A or B) 5 P(A) 1 P(B) 2 P(A and B)

0.82 5 0.32 1 P(B) 2 0.11

0.61 5 P(B)

P(B) 5 61%


0.6 5 P(A) 1 0.17 2 0.03

0.46 5 P(A)


0.5 5 0.2 1 0.4 2 P(A and B)

P(A and B) 5 0.1

46. d 5 5, a1 5 2

an 5 a1 1 (n 2 1)d

an 5 2 1 (n 2 1)(5) 5 23 1 5n

47. d 5 8, a1 5 227

an 5 a1 1 (n 2 1)d

an 5 227 1 (n 2 1)(8) 5 235 1 8n

48. d 5 218, a8 5 72

an 5 a1 1 (n 2 1)d

a8 5 a1 1 (8 2 1)d

72 5 a1 1 7(218)

198 5 a1

an 5 198 1 (n 2 1)(218) 5 216 2 18n

49. d 5 27, a7 5 28

an 5 a1 1 (n 2 1)d

a7 5 a1 1 (7 2 1)d

28 5 a1 1 6(27)

34 5 a1

an 5 34 1 (n 2 1)(27) 5 41 2 7n

50. d 5 6.5, a5 5 92

an 5 a1 1 (n 2 1)d

a5 5 a1 1 (5 2 1)d

92 5 a1 1 4(6.5)

66 5 a1

an 5 66 1 (n 2 1)(6.5) 5 59.5 1 6.5n

51. d 5 21.5, a9 5 4

an 5 a1 1 (n 2 1)d

a9 5 a1 1 (9 2 1)d

4 5 a1 1 8(21.5)

16 5 a1

an 5 16 1 (n 2 1)(21.5) 5 17.5 2 1.5n

52. r 5 2.5, a3 5 25 53. r 5 23, a2 5 218

an 5 a1rn 2 1 an 5 a1rn 2 1

a3 5 a1r3 2 1 a2 5 a1r2 2 1

25 5 a1(2.5)2 218 5 a1(23)1

4 5 a1 6 5 a1

an 5 4(2.5)n 2 1 an 5 6(23)n 2 1

54. r 5 20.25, a5 5 40.5

an 5 a1rn 2 1

a5 5 a1r5 2 1

240.5 5 a1(0.25)4

210,368 5 a1

an 5 210,368(0.25)n 2 1


n2ws-1200-b.indd 694 6/27/06 11:35:50 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.

695Algebra 2


55. r 5 6, a4 5 24 56. r 5 0.75, a6 5 30

an 5 a1rn 2 1 an 5 a1r n 2 1

a4 5 a1r4 2 1 a6 5 a1r6 2 1

24 5 a1(6)3 30 5 a1(0.75)5

1 }

9 5 a1

10,240 }

81 5 a1

an 5 1 } 9 (6)n 2 1 an 5

10,240 } 81 (0.75)n 2 1

57. r 5 4, a4 5 128

an 5 a1rn 2 1

a4 5 a1r4 2 1

128 5 a1(4)3

2 5 a1

an 5 2(4)n 2 1

Lesson 12.5

Investigating Algebra Activity 12.5 (p. 826)

1. Sequence: 4; 28; 196; 1372; 9604; 67,228; 470,596; 3,294,172


2. a1 5 15, d 5 11 2 15 5 24

an 5 an 2 1 2 4

3. a1 5 81, r 5 27

} 81 5 1 } 3

an 5 1 } 3 an 2 1

4. Arithmetic sequences: an 5 an 2 1 1 d

Geometric sequences: an 5 ran 2 1


1. a1 5 3

a2 5 a1 2 7 5 3 2 7 5 24

a3 5 a2 2 7 5 24 2 7 5 211

a4 5 a3 2 7 5 211 2 7 5 218

a5 5 a4 2 7 5 218 2 7 5 225

2. a0 5 162

a1 5 0.5a0 5 0.5(162) 5 81

a2 5 0.5a1 5 0.5(81) 5 40.5

a3 5 0.5a2 5 0.5(40.5) 5 20.25

a4 5 0.5a3 5 0.5(20.25) 5 10.125

3. a0 5 1

a1 5 a0 1 1 5 1 1 1 5 2

a2 5 a1 1 2 5 2 1 2 5 4

a3 5 a2 1 3 5 4 1 3 5 7

a4 5 a3 1 4 5 7 1 4 5 11

4. a1 5 4

a2 5 2a1 2 1 5 2(4) 2 1 5 7

a3 5 2a2 2 1 5 2(7) 2 1 5 13

a4 5 2a3 2 1 5 2(13) 2 1 5 25

a5 5 2a4 2 1 5 2(25) 2 1 5 49

5. a1 5 21, r 5 14

} 2 5 7

an 5 r p an 2 1

an 5 7an 2 1

6. a1 5 19, d 5 13 2 19 5 26

an 5 an 2 1 1 d

an 5 an 2 1 2 6

7. a1 5 11, d 5 22 2 11 5 11

an 5 an 2 1 1 d

an 5 an 2 1 1 11

8. a1 5 324, r 5 108

} 324 5 1 } 3

an 5 r p an 2 1

an 5 1 } 3 an 2 1

9. a1 5 1

a2 5 2

a3 5 2 5 a1 p a2

a4 5 4 5 a2 p a3

a1 5 1, a2 5 2, an 5 (an 2 2)(an 2 1)

10. a1 5 50,000, an 5 0.7an 2 1 1 5000

The number of members stabilizes at about 16,667 members.

11. f (x) 5 4x 2 3, x0 5 2

x1 5 f (x0) x2 5 f (x1) x3 5 f (x2)

5 f (2) 5 f (5) 5 f (17)

5 4(2) 2 3 5 4(5) 2 3 5 4(17) 2 3

5 5 5 17 5 65

12. f (x) 5 x2 2 5, x0 5 21

x1 5 f (x0) x2 5 f (x1) x3 5 f (x2)

5 f (21) 5 f (24) 5 f (11)

5 (21)2 2 5 5 (24)2 2 5 5 (11)2 2 5

5 24 5 11 5 116

12.5 Exercises (pp. 830–833)

Skill Practice

1. The repeated composition of a function with itself is called iteration.

2. An explicit rule gives the value based on the position of the term in the sequence. A recursive rule gives the value based on the previous term in the sequence.


n2ws-1200-b.indd 695 6/27/06 11:35:54 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.


3. a1 5 1

a2 5 a1 1 3 5 1 1 3 5 4

a3 5 a2 1 3 5 4 1 3 5 7

a4 5 a3 1 3 5 7 1 3 5 10

a5 5 a4 1 3 5 10 1 3 5 13

4. a0 5 4

a1 5 2a0 5 2(4) 5 8

a2 5 2a1 5 2(8) 5 16

a3 5 2a2 5 2(16) 5 32

a4 5 2a3 5 2(32) 5 64

5. a1 5 21

a2 5 a1 2 5 5 21 2 5 5 26

a3 5 a2 2 5 5 26 2 5 5 211

a4 5 a3 2 5 5 211 2 5 5 216

a5 5 a4 2 5 5 216 2 5 5 221

6. a0 5 3

a1 5 a0 2 12 5 3 2 1 5 2

a2 5 a1 2 22 5 2 2 4 5 22

a3 5 a2 2 32 5 22 2 9 5 211

a4 5 a3 2 42 5 211 2 16 5 227

7. a1 5 2

a2 5 a12 1 1 5 22 1 1 5 5

a3 5 a22 1 1 5 52 1 1 5 26

a4 5 a32 1 1 5 262 1 1 5 677

a5 5 a42 1 1 5 6772 1 1 5 458,330

8. a0 5 4

a1 5 (a0)2 2 10 5 42 2 10 5 6

a2 5 (a1)2 2 10 5 62 2 10 5 26

a3 5 (a2)2 2 10 5 262 2 10 5 666

a4 5 (a3)2 2 10 5 6662 2 10 5 443,546

9. a1 5 2

a2 5 22 1 3(2) 2 a1 5 4 1 6 2 2 5 8

a3 5 32 1 3(3) 2 a2 5 9 1 9 2 8 5 10

a4 5 42 1 3(4) 2 a3 5 16 1 12 2 10 5 18

a5 5 52 1 3(5) 2 a4 5 25 1 15 2 18 5 22

10. a0 5 2

a1 5 4

a2 5 a1 2 a0 5 4 2 2 5 2

a3 5 a2 2 a1 5 2 2 4 5 22

a4 5 a3 2 a2 5 22 2 2 5 24

11. a1 5 2

a2 5 3

a3 5 a2 p a1 5 3 p 2 5 6

a4 5 a3 p a2 5 6 p 3 5 18

a5 5 a4 p a3 5 18 p 6 5 108

12. A;

a1 5 1

a2 5 4

a3 5 a2 p a1 5 4 p 1 5 4

a4 5 a3 p a2 5 4 p 4 5 16

13. a1 5 21, d 5 14 2 21 5 27

an 5 an 2 1 1 d 5 an 2 1 2 7

14. a1 5 3, r 5 12

} 3 5 4

an 5 ran 2 1 5 4an 2 1

15. a1 5 4, r 5 2 12

} 4 5 23

an 5 ran 2 1 5 23an 2 1

16. a1 5 1, d 5 8 2 1 5 7

an 5 an 2 1 1 d 5 an 2 1 1 7

17. a1 5 44, r 5 11

} 44 5 1 } 4

an 5 ran 2 1 5 1 } 4 an 2 1

18. a1 5 1, a2 5 4

a3 5 5 5 a2 1 a1

a4 5 9 5 a3 1 a2

a1 5 1, a2 5 4, an 5 an 2 1 1 an 2 2

19. a1 5 54, d 5 43 2 54 5 211

an 5 an 2 1 1 d 5 an 2 1 2 11

20. a1 5 3, a2 5 5

a3 5 15 5 a2 p a1

a4 5 75 5 a3 p a2

a1 5 3, a2 5 5, an 5 an 2 1 p an 2 2

21. a1 5 16, a2 5 9

a3 5 7 5 a1 2 a2

a4 5 2 5 a2 2 a3

a1 5 16, a2 5 9, an 5 an 2 2 2 an 2 1

22. The previous terms must be defi ned fi rst.

a1 5 5, a2 5 2, an 5 an 2 2 2 an 2 1

23. The rule does not work for all the terms in the sequence.

a1 5 5, a2 5 2, an 5 an 2 2 2 an 2 1

24. f (x) 5 3x 2 2, x0 5 2

x1 5 f (x0) x2 5 f (x1) x3 5 f (x2)

5 f (2) 5 f (4) 5 f (10)

5 3(2) 2 2 5 3(4) 2 2 5 3(10) 2 2

5 4 5 10 5 28

25. f (x) 5 5x 1 6, x0 5 22

x1 5 f (x0) x2 5 f (x1) x3 5 f (x2)

5 f (22) 5 f (24) 5 f (214)

5 5(22) 1 6 5 5(24) 1 6 5 5(214) 1 6

5 24 5 214 5 264


n2ws-1200-b.indd 696 6/27/06 11:35:59 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.

697Algebra 2


26. g(x) 5 24x 1 7, x0 5 1

x1 5 g(x0) x2 5 g(x1) x3 5 g(x2)

5 g(1) 5 g(3) 5 g(25)

5 24(1) 1 7 5 24(3) 1 7 5 24(25) 1 7

5 3 5 25 5 27

27. f (x) 5 1 } 2 x 2 3, x0 5 2

x1 5 f (x0) x2 5 f (x1) x3 5 f (x2)

5 f (2) 5 f (22) 5 f (24)

5 1 } 2 (2) 2 3 5

1 } 2 (22) 2 3 5

1 } 2 (24) 2 3

5 22 5 24 5 25

28. f (x) 5 2 } 3 x 1 5, x0 5 6

x1 5 f (x0) x2 5 f (x1) x3 5 f (x2)

5 f (6) 5 f (9) 5 f (11)

5 2 } 3 (6) 1 5 5

2 } 3 (9) 1 5 5

2 } 3 (11) 1 5

5 9 5 11 5 12 1 }

3

29. h(x) 5 x2 2 4, x0 5 23

x1 5 h(x0) x2 5 h(x1) x3 5 h(x2)

5 h(23) 5 h(5) 5 h(21)

5 (23)2 2 4 5 (5)2 2 4 5 (21)2 2 4

5 5 5 21 5 437

30. f (x) 5 2x2 1 1, x0 5 21

x1 5 f (x0) x2 5 f(x1) x3 5 f (x2)

5 f (21) 5 f (3) 5 f (19)

5 2(21)2 1 1 5 2(3)2 1 1 5 2(19)2 1 1

5 3 5 19 5 723

31. f (x) 5 x2 2 x 1 2, x0 5 1

x1 5 f (x0) x2 5 f(x1) x3 5 f (x2)

5 f (1) 5 f (2) 5 f (4)

5 12 2 1 1 2 5 22 2 2 1 2 5 42 2 4 1 2

5 2 5 4 5 14

32. g(x) 5 23x2 1 2x, x0 5 2

x1 5 g(x0) x2 5 g(x1)

5 g(2) 5 g(28)

5 23(2)2 1 2(2) 5 23(28)2 1 2(28)

5 28 5 2208

x3 5 g(x2)

5 g(2208)

5 23(2208)2 1 2(2208)

5 2130,208

33. C;

f (x) 5 22x2 1 3, x0 5 2

x1 5 f (x0) x2 5 f(x1) x3 5 f (x2)

5 f (2) 5 f (21) 5 f (5)

5 22(2) 1 3 5 22(21)2 1 3 5 22(5) 1 3

5 21 5 5 5 27

34. a1 5 3, a2 5 8

a3 5 17 5 a12 1 a2

a4 5 81 5 a22 1 a1

a5 5 370 5 a32 1 a4

a1 5 3, a2 5 8, an 5 an 2 22 1 an 2 1

35. a1 5 1, a2 5 2

a3 5 12 5 4(a1 1 a2)

a4 5 56 5 4(a2 1 a3)

a5 5 272 5 4(a3 1 a4)

a1 5 1, a2 5 2, an 5 4(an 2 2 1 an 2 1)

36. a1 5 5

a2 5 5 Ï}

3 5 Ï}

3 a1

a3 5 15 5 Ï}

3 a2

a4 5 15 Ï}

3 5 Ï}

3 a3

a5 5 45 5 Ï}

3 a4

a1 5 5, an 5 Ï}

3 an 2 1

37. a1 5 2, a2 5 5

a3 5 11 5 a2 1 3a1

a4 5 26 5 a3 1 3a2

a5 5 59 5 a4 1 3a3

a1 5 2, a2 5 5, an 5 an 2 1 1 3an 2 2

38. a1 5 8, a2 5 4

a3 5 2 5 a1

} a2

a4 5 2 5 a2

} a3

a5 5 1 5 a3

} a4

a1 5 8, a2 5 4, an 5 an 2 2

} an 2 1

39. a1 5 23, a2 5 22

a3 5 5 5 2(a2 1 a1)

a4 5 23 5 2(a3 1 a2)

a5 5 22 5 2(a4 1 a3)

a1 5 3, a2 5 22, an 5 2(an 2 1 1 an 2 2)

40. Sample answer:

a1 5 1, a2 5 2, a3 5 3, an 5 an 2 3 1 an 2 2 1 an 2 1

First eight terms: 1, 2, 3, 6, 11, 20, 37, 68

41. Sample answer:

Because x1 5 2 and x2 5 f (x1) 5 f (2) 5 2, then x3 5 f (x2) 5 f (2) 5 2, x4 5 f (x3) 5 f (2) 5 2, and so on. So, if there is a function f and an initial value x0 such that the fi rst two iterates are equal, then all of the iterates must be equal.


n2ws-1200-b.indd 697 6/27/06 11:36:03 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.


42. a. a1 5 5

a2 5 3a1 1 3 5 3(5) 1 3 5 18

a3 5 a2

} 2 5 18

} 2 5 9

a4 5 3a3 1 3 5 3(9) 1 3 5 30

a5 5 a4

} 2 5 30

} 2 5 15

a6 5 3a5 1 3 5 3(15) 1 3 5 48

a7 5 a6

} 2 5 48

} 2 5 24

a8 5 a7

} 2 5 24

} 2 5 12

a9 5 a8

} 2 5 12

} 2 5 6

a10 5 a9

} 2 5 6 } 2 5 3

b. Sample answer:

If a1 5 1, the fi rst ten terms are 1, 6, 3, 12, 6, 3, 12, 6, 3, 12.



The terms of the sequence will always eventually repeat the numbers 3, 12, 6.

Problem Solving

43. a. Fish at start of year n 5 0.8 p

Fish at start of year n 2 1

1 New fi sh added

an 5 0.8an 2 1 1 500

a1 5 5000, an 5 0.8an 2 1 1 500

The fi rst fi ve terms are 5000, 4500, 4100, 3780, 3524.

At the beginning of the fi fth year, there will be 3524 fi sh in the lake.

b. Over time, the population approaches 2500 fi sh.

44.

Amount of chlorine at start of week n

5 0.6 p

Amount of chlorine at start of week n 2 1

1 New chlorine added

an 5 0.6 p an 2 1 1 16

a1 5 34, an 5 0.6an 2 1 1 16

Over time, the amount of chlorine in the pool approaches 40 ounces.

45. Current balance 5 1.014 p

Previous balance

2 Payment

an 5 1.014 p an 2 1 2 100

a1 5 2000, an 5 1.014an 2 1 2 100

It will take Gladys 24 months to pay off her credit card bill. Because a24 5 62.14, the balance at the beginning of the 24th month is $62.14. So, she will be able to pay off the balance at the end of the 24th month.

46. f1 5

1 }

Ï}

5 1 1 1 Ï

} 5 }

2 2 1 2

1 }

Ï}

5 1 1 2 Ï

} 5 }

2 2 1 5 1

f2 5 1 }

Ï}

5 1 1 1 Ï

} 5 }

2 2 2 2

1 }

Ï}

5 1 1 2 Ï

} 5 }

2 2 2 5 1

f3 5 1 }

Ï}

5 1 1 1 Ï

} 5 }

2 2 3 2

1 }

Ï}

5 1 1 2 Ï

} 5 }

2 2 3 5 2

f4 5 1 }

Ï}

5 1 1 1 Ï

} 5 }

2 2 4 2

1 }

Ï}

5 1 1 2 Ï

} 5 }

2 2 4 5 3

f5 5 1 }

Ï}

5 1 1 1 Ï

} 5 }

2 2 5 2

1 }

Ï}

5 1 1 2 Ï

} 5 }

2 2 5 5 5

47. a. Current amount 5 0.70 p

Previous amount

1 New dose

an 5 0.70 p an 2 1 1 20

a1 5 20, an 5 0.70an 2 1 1 20

b. The maintenance level of the drug is 66 2 }

3 milligrams.

c. The new recursive rule would be

a1 5 2(20), an 5 0.7an 2 1 1 2(20), or

a1 5 40, an 5 0.7an 2 1 1 40.

The new maintenance level would be doubled as well,

to 133 1 }

3 milligrams.

48. a. Current balance 5 1.08 p

Previous balance

2 Amount withdrawn

an 5 1.08 p an 2 1 2 30,000

b. an 5 1.08an 2 1 2 30,000

an 1 30,000 5 1.08an 2 1

an 1 30,000

} 1.08

5 an 2 1

If a20 5 0, then a19 5 0 1 30,000

} 1.08

a18 5 a19 1 30,000

} 1.08

a0 ø 294,544.42 (use calculator).

You should have at least $294,544.42 in your account when you retire.

Mixed Review

49. a2 1 b2 5 c2 50. a2 1 b2 5 c2

32 1 32 5 x2 52 1 x2 5 92

Ï}

18 5 x x2 5 Ï}

56

3 Ï}

2 5 x x 5 2 Ï}

14

51. a2 1 b2 5 c2

122 1 x2 5 152

x2 5 Ï}

81

x 5 9

52. 163/2 5 1 2 Ï}

16 2 3 5 43 5 64

53. (2243)2/5 5 1 5 Ï}

2243 2 2 5 (23)2 5 9

54. 6421/2 5 1 }

641/2 5 1 }

2 Ï}

64 5

1 } 8


n2ws-1200-b.indd 698 6/27/06 11:36:08 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.

699Algebra 2


Quiz 12.4–12.5 (p. 833)

1. a1 5 2, r 5 3 } 7 2. a1 5 4, r 5 2

5 } 6

S 5 a1 } 1 2 r S 5

a1 } 1 2 r

S 5 2 }

1 2 3 } 7 5

7 } 2 S 5

4 }

1 2 1 2 5 } 6 2

5 24

} 11

3. r 5 15

} 8 }

3 }

4 5

5 } 2

Because 5 }


4. 0.777. . . 5 7(0.1) 1 7(0.1)2 1 7(0.1)3 1 . . .

5 a1 } 1 2 r

5 7(0.1)

} 1 2 0.1

5 0.7

} 0.9

5 7 } 9

5. 0.393939. . . 5 39(0.01) 1 39(0.01)2 1 39(0.01)3 1 . . .

5 a1 } 1 2 r

5 39(0.01)

} 1 2 0.01

5 0.39

} 0.99

5 13

} 33

6. 123.123123. . .

5 123 1 123(0.001) 1 123(0.001)2 1 . . .

5 a1 } 1 2 r

5 123 } 1 2 0.001

5 123

} 0.999

5 41,000

} 333

7. a1 5 2

a2 5 a1 1 4 5 2 1 4 5 6

a3 5 a2 1 4 5 6 1 4 5 10

a4 5 a3 1 4 5 10 1 4 = 14

a5 5 a4 1 4 5 14 1 4 5 18

8. a0 5 3

a1 5 (a0)2 2 5 5 32 2 5 5 4

a2 5 (a1)2 2 5 5 42 2 5 5 11

a3 5 (a2)2 2 5 5 112 2 5 = 116

a4 5 (a3)2 2 5 5 1162 2 5 5 13,451

9. a1 5 1

a2 5 4

a3 5 a2 2 a1 5 4 2 1 5 3

a4 5 a3 2 a2 5 3 2 4 5 21

a5 5 a4 2 a3 5 21 2 3 5 24

10. a1 5 5

a2 5 17

} 4 5 a1 2

3 } 4

a3 5 7 } 2 5 a2 2

3 } 4

a4 5 11

} 4 5 a3 2 3 } 4

a5 5 2 5 a4 2 3 } 4

a1 5 5, an 5 an 2 1 2 3 } 4

11. a1 5 2, a2 5 6

a3 5 12 5 a1 p a2

a4 5 72 5 a2 p a3

a5 5 864 5 a3 p a4

a1 5 2, a2 5 6, an 5 an 2 2 p an 2 1

12. a1 5 8

a2 5 24 5 3a1

a3 5 72 5 3a2

a4 5 216 5 3a3

a5 5 648 5 3a4

a1 5 8, an 5 3an 2 1

13. f (x) 5 23x 2 2, x0 5 1

x1 5 f (x0) x2 5 f (x1) x3 5 f (x2)

5 f (1) 5 f (25) 5 f (13)

5 23(1) 2 2 5 23(25) 2 2 5 23(13) 2 2

5 25 5 13 5 241

14. g(x) 5 4x 1 1, x0 5 2

x1 5 f (x0) x2 5 f (x1) x3 5 f (x2)

5 f (2) 5 f (9) 5 f (37)

5 4(2) 1 1 5 4(9) 1 1 5 4(37) 1 1

5 9 5 37 5 149

15. f (x) 5 22x 1 3, x0 5 22

x1 5 f (x0) x2 5 f (x1) x3 5 f (x2)

5 f (22) 5 f (7) 5 f (211)

5 22(22) 1 3 5 22(7) 1 3 5 22(211) 1 3

5 7 5 211 5 25

16. f (x) 5 5x 2 7, x0 5 23

x1 5 f (x0) x2 5 f (x1) x3 5 f (x2)

5 f (23) 5 f (222) 5 f (2117)

5 5(23) 2 7 5 5(222) 2 7 5 5(2117) 2 7

5 222 5 2117 5 2592


n2ws-1200-b.indd 699 6/27/06 11:36:13 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.


17. h(x) 5 x2 2 6, x0 5 21

x1 5 f (x0) x2 5 f (x1) x3 5 f (x2)

5 f (21) 5 f (25) 5 f (19)

5 (21)2 2 6 5 (25)2 2 6 5 (19)2 2 6

5 25 5 19 5 355

18. f (x) 5 3x2 1 2, x0 5 0

x1 5 f (x0) x2 5 f (x1) x3 5 f (x2)

5 f (0) 5 f (2) 5 f (14)

5 3(0)2 1 2 5 3(2)2 1 2 5 3(14)2 1 2

5 2 5 14 5 590

19. a1 5 25, r 5 0.85

S 5 a1 } 1 2 r

S 5 25 } 1 2 0.85 5 166

2 }

3

The pendulum swings 166 2 }

3 inches.

Problem Solving Workshop 12.5 (p. 835)

1. Let L be the limit of the sequence.

L 5 0.25L 1 300

0.75L 5 300

L 5 400

The sequence approaches 400.

2. Let L be the limit of the sequence.

L 5 0.38L 1 512

0.62L 5 512

L ø 825.8

The sequence approaches 825.8.

3. Graphing method:

Y= : nMin 5 1

u(n) 5 0.92u(n 2 1) 1 1200

u(n 2 Min) 5 50,000

WINDOW :

nMin 5 1 Xmin 5 0 Ymin 5 5000

nMax 5 1 Xmax 5 200 Ymax 5 35,000

PlotStart 5 1 Xscl 5 50 Yscl 5 5000

PlotStep 5 1

n=150X=150 Y=15000.141

Algebraic method:

a1 5 50,000, an 5 0.92an 2 1 1 1200

L 5 0.92L 1 1200

0.08L 5 1200

L 5 15,000

With each method, the number of members approaches 15,000.

4. Graphing method:

Y= : nMin 5 1

u(n) 5 0.98u(n 2 1) 1 1150

u(n 2 Min) 5 54,000

WINDOW :

nMin 5 1 Xmin 5 0 Ymin 5 45,000

nMax 5 600 Xmax 5 600 Ymax 5 60,000

PlotStart 5 1 Xscl 5 50 Yscl 5 5000

PlotStep 5 10

n=550X=550 Y=57499.947

Algebraic method:

a1 5 54,000, an 5 0.98an 2 1 1 1150

L 5 0.98L 1 1150

0.02L 5 1150

L 5 57,500

With each method, the number of books in the library approaches 57,500.

5. Each year, 2% of the books are lost or discarded, so 98% of the books remain. The coeffi cient of an 2 1 should be 0.98, not 0.02.

a1 5 54,000, an 5 0.98an 2 1 1 1150

Let L be the limit of the sequence. Then:

L 5 0.98L 1 1150

0.02L 5 1150

L 5 57,500

6. Sample answer: Save a penny on day one, and increase each day’s savings by a penny.

a1 5 0.01, an 5 an 2 1 1 0.01

12.5 Extension (p. 837)

1. Prove: i 5 1

∑ n

(2i 2 1) 5 n2

Basis Step: 2(1) 2 1 0 12 → 1 5 1 ✓

Inductive Step:

1 1 3 1 5 1 . . . 1 (2k 2 1) 5 k2

1 1 3 1 5 1 . . . 1 (2k 2 1) 1 (2(k 1 1) 2 1

5 k2 1 (2(k 1 1) 2 1)

5 k2 1 2k 1 1

5 (k 1 1)2

Therefore, i 5 1

∑ n

(2i 2 1) 5 n2 for all positive integers n.

2. Prove: i 5 1

∑ n

i2 5 n(n 1 1)(2n 1 1)

}} 6

Basis Step: 12 0 1(1 1 1)(2(1) 1 1)

}} 6 → 1 5 1 ✓


n2ws-1200-b.indd 700 6/27/06 11:36:22 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.

701Algebra 2


Inductive Step:

1 1 4 1 9 1 . . . 1 k2 5 k(k 1 1)(2k 1 1)

}} 6

1 1 4 1 9 1 . . . 1 k2 1 (k 1 1)2

5 k(k 1 1)(2k 1 1)

}} 6 1 (k 1 1)2

5 k(k 1 1)(2k 1 1) 1 6(k 1 1)2

}}} 6

5 (k 1 1)[k(2k 1 1) 1 6(k 1 1)]

}}} 6

5 (k 1 1)(2k2 1 7k 1 6)

}} 6

5 (k 1 1)(k 1 2)(2k 1 3)

}} 6

5 (k 1 1)((k 1 1) 1 1)((2k 1 2) 1 1)

}}} 6

Therefore, i 5 1

∑ n

i2 5 n(n 1 1)(2n 1 1)

}} 6 for all positive

integers n.

3. Prove: i 5 1

∑ n

2i 2 1 5 2n 2 1

Basis Step: 21 2 1 0 21 2 1 → 1 5 1 ✓

Inductive Step:

1 1 2 1 4 1 . . . 1 2k 2 1 5 2k 2 1

1 1 2 1 4 1 . . . 1 2k 2 1 1 2(k 1 1) 2 1

5 2k 2 1 1 2(k 1 1) 2 1

5 2(2k) 2 1

5 2k 1 1 2 1

Therefore, i 5 1

∑ n

2i 2 1 5 2n 2 1 for all positive integers n.

4. Prove: i 5 1

∑ n

a1r i 2 1 5 a1 1 1 2 rn }

1 2 r 2

Basis Step: a1r 1 2 1 0 a1 1 1 2 r1 }

1 2 r 2 → a1 5 a1 ✓

Inductive Step:

a1 1 a1r 1 a1r2 1 . . . 1 a1rk 2 1 5 a1 1 1 2 r k }

1 2 r 2

a1 1 a1r 1 a1r2 1 . . . 1 a1rk 2 1 1 a1r (k 1 1) 2 1

5 a1 1 1 2 r k }

1 2 r 2 1 a1r(k 1 1) 2 1

5 a1 1 1 2 rk }

1 2 r 1 rk 2

5 a1 1 1 2 rk 1 1 }

1 2 r 2

Therefore,

i 5 1 ∑

n

a1r i2 1 5 a1 1 1 2 rn }

1 2 r 2 for all positive

integers n.

5. Prove: i 5 1

∑ n

1 }

i(i 1 1) 5

n } n 1 1

Basis Step: 1 }

1(1 1 1) 0

1 }

1 1 1 →

1 }

2 5

1 } 2 ✓

Inductive Step:

1 }

2 1

1 } 6 1

1 } 12 1 . . . 1

1 }

k(k 1 1) 5

k }

k 1 1

1 }

2 1

1 } 6 1

1 } 12 1 . . . 1

1 }

k(k 1 1) 1

1 }}

(k 1 1)(k 1 2)

5 k }

k 1 1 1

1 }}

(k 1 1)(k 1 2)

5 k(k 1 2) 1 1

}} (k 1 1)(k 1 2)

5 (k 1 1)2

}} (k 1 1)(k 1 2)

5 k 1 1

} k 1 2

5 k 1 1 }

(k 1 1) 1 1

Therefore i 5 1

∑ n

1 }

i(i 1 1) 5

n } n 1 1 for all positive integers n.

6. Prove: i 5 1

∑ n

(2i)2 5 2n(n 1 1)(2n 1 1)

}} 3

Basis Step: (2 p 1)2 0 2 p 1(1 1 1)(2 p 1 1 1)

}} 3 → 4 5 4 ✓

Inductive Step:

4 1 16 1 36 1 . . . 1 (2k)2 5 2k(k 1 1)(2k 1 1)

}} 3

4 1 16 1 36 1 . . . 1 (2k)2 1 (2(k 1 1))2

5 2k(k 1 1)(2k 1 1)

}} 3 1 (2(k 1 1))2

5 2k(k 1 1)(2k 1 1)

}} 3 1 4(k 1 1)2

5 2k(k 1 1)(2k 1 1) 1 12(k 1 1)2

}}} 3

5 2(k 1 1)[k(2k 1 1) 1 6(k 1 1)]

}}} 3

5 2(k 1 1)(k 1 2)(2k 1 3)

}} 3

5 2(k 1 1)[(k 1 1) 1 1][2(k 1 1) 1 1]

}}} 3

Therefore, i 5 1

∑ n

(2i)2 5 2n(n 1 1)(2n 1 1)

}} 3 for all positive

integers n.

7. Series: 1, 6, 15, 28. . .

A recursive formula for the nth hexagonal number is Hn 5 Hn 2 1 1 4n 2 3.

Prove: i 5 1

∑ n

4i 2 3 5 n(2n 2 1)

Basis Step: 4 p 1 2 3 0 1(2 p 1 2 1) → 1 5 1 ✓


n2ws-1200-b.indd 701 6/27/06 11:36:26 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.


Inductive Step:

1 1 5 1 9 1 . . . 1 4k 2 3 5 k(2k 2 1)

1 1 5 1 9 1 . . . 1 4k 2 3 1 4(k 1 1) 2 3

5 k(2k 2 1) 1 4(k 1 1) 2 3

5 2k2 1 3k 1 1

5 (k 1 1)(2k 1 1)

5 (k 1 1)[2(k 1 1) 2 1]

Therefore, the nth hexagonal number is given by n(2n 2 1) for all positive integers n.

8. Prove: f1 1 f2 1 f3 1 . . . 1 fn 5 fn 1 2 2 1

Basis Step: f1 0 f3 2 1 → 1 0 2 2 1 → 1 5 1 ✓

Inductive Step:

f1 1 f2 1 . . . 1 fk 5 fk 1 2 2 1

f1 1 f2 1 . . . 1 fk 1 fk 1 1 5 fk 1 2 1 fk 1 1 2 1

5 fk 1 3 2 1

5 f(k 1 1) 1 2 2 1

Therefore, the sum of the first n Fibonacci numbers is fn 1 2 2 1 for all postive integers n.

Mixed Review of Problem Solving (p. 838)

1. a. a1 5 2(12(0.7)) 5 16.8, r 5 0.7

an 5 a1rn 2 1

an 5 16.8(0.7)n 2 1

Distance 5 i 5 1

∑ `

16.8(0.7)i 2 1

b. S 5 a1 } 1 2 r

S 5 16.8

} 1 2 0.7 5 56

The ball travels a total of 56 1 12 5 68 feet.

2. a. Number of new branches: 1, 2, 4, 8, 16, 32

b. The sequence is geometric with a common ratio of 2.

c. Explicit rule: a1 5 1, r 5 2 } 1 5 2

an 5 a1rn 2 1

an 5 2n 2 1

Recursive rule: a1 5 1, r 5 2

an 5 ran 2 1

a1 5 1, an 5 2an 2 1

3. f (x) 5 x2 2 8, x0 5 2

x1 5 f (x0) x2 5 f (x1) x3 5 f (x2)

5 f (2) 5 f (24) 5 f (8)

5 22 2 8 5 (24)2 2 8 5 82 2 8

5 24 5 8 5 56

x1 1 x2 1 x3 5 24 1 8 1 56 5 60

4. Sample answer: Let a1 5 4 and d 5 3.

Explicit rule: an 5 a1 1 (n 2 1)d

an 5 4 1 (n 2 1)(3) 5 1 1 3n

Recursive rule: an 5 an 2 1 1 d

a1 5 4, an 5 an 2 1 1 3

5. Sample answer:

If 21 < r < 1, then rn gets closer and closer to zero as n increases, and the sum of the series approaches the

value a1 }

1 2 r .

Sn 5 a1 1 1 2 rn }

1 2 r 2 ø a1 1 1 2 0

} 1 2 r

2 5 a1 } 1 2 r

If r < 21 or r > 1, then rn gets further away from zero as n increases, and the sum does not approach a certain number.

If r 5 1, the sum does not exist because the value is undefi ned.

Sn 5 a1 1 1 2 rn }

1 2 r 2 5 a1 1 1 2 1n

} 1 2 1

2 5 a1 1 0 } 0 2 Undefi ned

6. The length would be fi nite because the common ratio is 0.9, which is less than 1.

a1 5 16, r 5 0.9

S 5 a1 } 1 2 r

S 5 16 } 1 2 0.9 5 160

The length of the spring would be 160 inches.

7. a. 6.5

} 12

ø 0.542

The monthly interest rate is approximately 0.542%.

Amount owed 5

(1 1 interest)(Current balance) 2 Payment

Recursive rule: a1 5 10,000, an 5 1.00542an 2 1 2 196

b. After 12 months, you will owe about $8246.37.

c. a1 5 10,000; an 5 1.00542an 2 1 2 246

If you pay $246 a month, the loan will be repaid in 47 months.

d. Sample answer:

Yes, it is benefi cial to pay the extra $50 each month. This causes the balance to decrease faster, so you don’t have to pay as much interest.

8. Number of trees

5 0.9 p Previous number of trees

1 New trees

an 5 0.9an 2 1 1 500

a1 5 8000, an 5 0.9an 2 1 1 500

Over an extended period of time, there will be about 5000 trees on the farm.


n2ws-1200-b.indd 702 6/27/06 11:36:30 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.

703Algebra 2


9. Sample answer:

S 5 a1 } 1 2 r

Let S 5 4 and choose r 5 1 }

2 .

4 5 a1 }

1 2 1 } 2

2 5 a1

So, i 5 1

∑ `

2 1 1 } 2 2 i 2 1

is an infi nite geometric series with a

sum of 4.

Chapter 12 Review (pp. 840–842)

1. The values in the range of a sequence are called the terms of the sequence.

2. A sequence is arithmetic if the difference between consecutive terms is constant.

3. An explicit rule gives an as a function of the term’s position number n in the sequence.

4. In a geometric sequence, the ratio of any term to the previous term is constant.

5. n 5 1

∑ 6

(n2 1 7) 5 (12 1 7) 1 (22 1 7) 1 (32 1 7) 1 (42 1 7) 1 (52 1 7) 1 (62 1 7) 5 8 1 11 1 16 1 23 1 32 1 43

5 133

6. i 5 2

∑ 6

(10 2 4i) 5 (10 2 4 p 2) 1 (10 2 4 p 3)

1 (10 2 4 p 4) 1 (10 2 4 p 5)

1 (10 2 4 p 6)

5 2 1 (22) 1 (26) 1 (210) 1 (214)

5 230

7. i 5 1

∑ 17

i 5 n(n 1 1)

} 2 (special formula)

5 17(17 1 1)

} 2

5 153

8. k 5 1

∑ 25

k2 5 n(n 1 1)(2n 1 1)

}} 6 (special formula)

5 25(25 1 1)(2 p 25 1 1)

}} 6

5 5525

9. a1 5 8, d 5 5 2 8 5 23

an 5 a1 1 (n 2 1)d

an 5 8 1 (n 2 1)(23)

an 5 11 2 3n

10. a8 5 54, d 5 7

an 5 a1 1 (n 2 1)d

a8 5 a1 1 (8 2 1)d

54 5 a1 1 7(7)

5 5 a1

an 5 5 1 (n 2 1)(7)

an 5 22 1 7n

11. a4 5 27, a11 5 69

a11 5 a1 1 (11 2 1)d → 69 5 a1 1 10d

a4 5 a1 1 (4 2 1)d → 27 5 a1 1 3d

42 5 7d

6 5 d

69 5 a1 1 10(6) → a1 5 9

an 5 a1 1 (n 2 1)d

an 5 9 1 (n 2 1)(6)

an 5 3 1 6n

12. a1 5 3 1 2(1) 5 5, a15 5 3 1 2(15) 5 33

Sn 5 n 1 a1 1 an }

2 2

S15 5 15 1 a1 1 a15 }

2 2

S15 5 15 1 5 1 33 }

2 2 5 285

13. a1 5 25 2 3(1) 5 22, a26 5 25 2 3(26) 5 253

Sn 5 n 1 a1 1 an }

2 2

S26 5 26 1 a1 1 a26 }

2 2

S26 5 26 1 22 1 (253) }

2 2 5 2403

14. a1 5 6(1) 2 5 5 1, a22 5 6(22) 2 5 5 127

Sn 5 n 1 a1 1 an }

2 2

S22 5 22 1 a1 1 a22 }

2 2

S22 5 22 1 1 1 127 }

2 2 5 1408

15. a1 5 284 1 8(1) 5 276, a30 5 284 1 8(30) 5 156

Sn 5 n 1 a1 1 an }

2 2

S30 5 30 1 a1 1 a30 }

2 2

S30 5 30 1 276 1 156 }

2 2 5 1200


n2ws-1200-b.indd 703 6/27/06 11:36:34 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.


16. a1 5 200 1 25 5 225, d 5 25

an 5 a1 1 (n 2 1)d

an 5 225 1 (n 2 1)(25)

an 5 200 1 25n

After n months, Joe will have paid 200 1 25n dollars toward the computer.

17. a1 5 256, r 5 64

} 256 5 1 } 4

an 5 a1rn 2 1

an 5 256 1 1 } 4 2 n 2 1

18. a2 5 200, r 5 5

an 5 a1rn 2 1

a2 5 a1r2 2 1

200 5 a1(5)

40 5 a1

an 5 40(5)n 2 1

19. a1 5 144, a3 5 16

an 5 a1rn 2 1

a3 5 a1r3 2 1

16 5 144r2

1 }

9 5 r2

6 1 } 3 5 r

an 5 144 1 1 } 3 2

n 2 1 or an 5 144 1 2

1 } 3 2 n 2 1

20. a1 5 3, r 5 5

S6 5 a1 1 1 2 r6 }

1 2 r 2

S6 5 3 1 1 2 56 }

1 2 5 2 5 11,718

21. a1 5 8, r 5 2

S9 5 a1 1 1 2 r9 }

1 2 r 2

S9 5 8 1 1 2 29 }

1 2 2 2 5 4088

22. a1 5 15, r 5 2 } 3

S5 5 a1 1 1 2 r5 }

1 2 r 2

S5 5 15 1 1 2 1 2 } 3 2 5 }

1 2 2 } 3 2 5

1055 } 27

23. a1 5 40, r 5 1 } 2

S7 5 a1 1 1 2 r7 }

1 2 r 2

S7 5 40 1 1 2 1 1 } 2 2 7 }

1 2 1 } 2 2 5

635 } 8

24. a1 5 3, r 5 5 } 8

S 5 a1 } 1 2 r

S 5 3 }

1 2 5 } 8 5 8

25. a1 5 7, r 5 2 3 } 4

S 5 a1 } 1 2 r

S 5 7 }

1 2 1 2 3 } 4 2 5 4

26. r 5 1.3

Because r ≥ 1, the series has no sum.

27. a1 5 20.2, r 5 0.5

S 5 a1 } 1 2 r

S 5 20.2

} 1 2 0.5 5 0.4

28. 0.888. . . 5 8(0.1) 1 8(0.1)2 1 8(0.1)3. . .

5 a1 } 1 2 r

5 8(0.1)

} 1 2 0.1

5 0.8

} 0.9

5 8 } 9

29. 0.546546546. . . 5

546(0.001) 1 546(0.001)2 1 546(0.001)3. . .

5 a1 } 1 2 r

5 546(0.001)

} 1 2 0.001

5 0.546

} 0.999

5 182

} 333

30. 0.3787878. . . 5 0.37 1 87(0.01)2 1 87(0.01)3 1 . . .

5 37

} 100 1 a1 } 1 2 r

5 37

} 100 1 87(0.01)2

} 1 2 0.01

5 37

} 100 1 0.0087

} 0.99

5 37

} 100 1 87 } 9900

5 25

} 66


n2ws-1200-b.indd 704 6/27/06 11:36:38 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.

705Algebra 2


31. 0.783838. . . 5 0.78 1 38(0.01)2 1 38(0.01)3 1 . . .

5 78

} 100 1 a1 } 1 2 r

5 78

} 100 1 38(0.01)2

} 1 2 0.01

5 78

} 100 1 0.0038

} 0.99

5 78

} 100 1 38 } 9900

5 388

} 495

32. a1 5 4

a2 5 a1 1 9 5 4 1 9 5 13

a3 5 a2 1 9 5 13 1 9 5 22

a4 5 a3 1 9 5 22 1 9 5 31

a5 5 a4 1 9 5 31 1 9 5 40

33. a1 5 8

a2 5 5a1 5 5(8) 5 40

a3 5 5a2 5 5(40) 5 200

a4 5 5a3 5 5(200) 5 1000

a5 5 5a4 5 5(1000) 5 5000

34. a1 5 2

a2 5 2a1 5 2(2) 5 4

a3 5 3a2 5 3(4) 5 12

a4 5 4a3 5 4(12) 5 48

a5 5 5a4 5 5(48) 5 240

35. a1 5 6, r 5 18

} 6 5 3

an 5 ran 2 1

a1 5 6, an 5 3an 2 1

36. The nth term is n greater than the previous term.

an 5 an 2 1 1 d

a1 5 4, an 5 an 2 1 1 n

37. a1 5 7, d 5 13 2 7 5 6

an 5 an 2 1 1 d

a1 5 7, an 5 an 2 1 1 6

38. Current population

5 1.01 p Previous population

an 5 1.01an 2 1

a1 5 26,000, an 5 1.01an 2 1

Chapter 12 Test (p. 843)

1. a2 2 a1 5 9 2 5 5 4

a3 2 a2 5 13 2 9 5 4

a4 2 a3 5 17 2 13 5 4

Arithmetic; the common difference is 4.

2. a2

} a1 5

6 } 3 5 2

a3

} a2 5

12 } 6 5 2

a4

} a3 5

24 } 12 5 2

Geometric; the common ratio is 2.

3. a2

} a1 5

10 } 40 5

1 } 4

a3

} a2 5

5 } 2 } 10 5

1 } 4

a4

} a3 5

5 } 8 }

5 }

2 5

1 } 4

Geometric; the common ratio is 1 }

4 .

4. Neither, there is no common ratio or common difference.

5. a1 5 6 2 12 5 5 6. a1 5 7 p 13 5 7

a2 5 6 2 22 5 2 a2 5 7 p 23 5 56

a3 5 6 2 32 5 23 a3 5 7 p 33 5 189

a4 5 6 2 42 5 210 a4 5 7 p 43 5 448

a5 5 6 2 52 5 219 a5 5 7 p 53 5 875

a6 5 6 2 62 5 230 a6 5 7 p 63 5 1512

7. a1 5 4

a2 5 5a1 5 5(4) 5 20

a3 5 5a2 5 5(20) 5 100

a4 5 5a3 5 5(100) 5 500

a5 5 5a4 5 5(500) 5 2500

a6 5 5a5 5 5(2500) 5 12,500

8. a1 5 21

a2 5 a1 1 6 5 21 1 6 5 5

a3 5 a2 1 6 5 5 1 6 5 11

a4 5 a3 1 6 5 11 1 6 5 17

a5 5 a4 1 6 5 17 1 6 5 23

a6 5 a5 1 6 5 23 1 6 5 29

9. Next term: 29

a1 5 5, d 5 11 2 5 5 6

an 5 a1 1 (n 2 1)d

an 5 5 1 (n 2 1)(6) 5 21 1 6n

10. Next term: 1875

a1 5 3, r 5 15

} 3 5 5

an 5 a1r n 2 1

an 5 3(5)n 2 1


n2ws-1200-b.indd 705 6/27/06 11:36:42 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.

706Algebra 2Worked-Out Solution Key706Algebra 2Worked-Out Solution Key


11. Next term: 10

} 25

a1 5 6 } 5 5

1 1 5 }

5(1)

a2 5 7 } 10 5

2 1 5 }

5(2)

a3 5 8 } 15 5

3 1 5 }

5(3)

an 5 n 1 5

} 5n

12. Next term: 8

a1 5 1.6 5 1.6(1)

a2 5 3.2 5 1.6(2)

a3 5 4.8 5 1.6(3)

an 5 1.6n

13. i 5 1

∑ 48

i 5 n(n 1 1)

} 2 5 48(48 1 1)

} 2 5 1176

14. n 5 1

∑ 28

n2 5 n(n 1 1)(2n 1 1)

}} 6 5 28(28 1 1)(2(28) 1 1)

}} 6 5 7714

15. a1 5 4(1) 2 9 5 25, a10 5 4(10) 2 9 5 31

Sn 5 n 1 a1 1 an }

2 2

S10 5 10 1 a1 1 a10 }

2 2

S10 5 10 1 25 1 31 }

2 2 5 130

16. a1 5 2(1) 1 5 5 7, a19 5 2(19) 1 5 5 43

Sn 5 n 1 a1 1 an }

2 2

S19 5 19 1 a1 1 a19 }

2 2

S19 5 19 1 7 1 43 }

2 2 5 475

17. a1 5 9, r 5 2

Sn 5 a1 1 1 2 rn }

1 2 r 2

S5 5 a1 1 1 2 r5 }

1 2 r 2

S5 5 9 1 1 2 25 }

1 2 2 2 5 279

18. a1 5 12, r 5 1 } 3

Sn 5 a1 1 1 2 rn }

1 2 r 2

S6 5 a1 1 1 2 r6 }

1 2 r 2

S6 5 12 1 1 2 1 1 } 3 2 6 }

1 2 1 1 } 3 2 2 5

1456 } 81

19. a1 5 8, r 5 3 } 4 20. a1 5 20, r 5

3 } 10

S 5 a1 } 1 2 r S 5

a1 } 1 2 r

S 5 8 }

1 2 1 3 } 4 2 5 32 S 5

20 }

1 2 1 3 } 10 2 5

200 } 7

21. 0.111. . . 5 1(0.1) 1 1(0.1)2 1 1(0.1)3 1 . . .

5 a1 } 1 2 r

5 1(0.1)

} 1 2 0.1

5 0.1

} 0.9

5 1 } 9

22. 0.464646. . . 5 46(0.01) 1 46(0.01)2 1 46(0.01)3 1 . . .

5 a1 } 1 2 r

5 46(0.01)

} 1 2 0.01

5 0.46

} 0.99

5 46

} 99

23. 0.187187187. . . 5 187(0.001) 1 187(0.001)2

1 187(0.001)3 1 . . .

5 a1 } 1 2 r

5 187(0.001)

} 1 2 0.001

5 0.187

} 0.999

5 187

} 999

24. 0.3252525. . . 5 0.32 1 52(0.01)2 1 52(0.01)3 1 . . .

5 32

} 100 1 a1 } 1 2 r

5 32

} 100 1 52(0.01)2

} 1 2 0.01

5 32

} 100 1 0.0052

} 0.99

5 32

} 100 1 52 } 9900

5 161

} 495

25. a1 5 2 26. a1 5 3

a2 5 12 5 6a1 a2 5 10 5 a1 1 7

a3 5 72 5 6a2 a3 5 17 5 a2 1 7

a4 5 432 5 6a3 a4 5 24 5 a3 1 7

a1 5 2, an 5 6an 2 1 a1 5 3, an 5 an 2 1 1 7


n2ws-1200-b.indd 706 6/27/06 11:36:47 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.

707Algebra 2


27. a1 5 135

a2 5 45 5 1 } 3 a1

a3 5 15 5 1 } 3 a2

a4 5 5 5 1 } 3 a3

a1 5 135, an 5 1 } 3 an 2 1

28. a1 5 1

a2 5 23 5 23a1

a3 5 9 5 23a2

a4 5 227 5 23a3

a1 5 1, an 5 23an 2 1

29. f (x) 5 3x 2 7, x0 5 4

x1 5 f (x0) x2 5 f (x1) x3 5 f (x2)

5 f (4) 5 f (5) 5 f (8)

5 3(4) 2 7 5 3(5) 2 7 5 3(8) 2 7

5 5 5 8 5 17

30. f (x) 5 8 2 5x, x0 5 1

x1 5 f (x0) x2 5 f (x1) x3 5 f (x2)

5 f (1) 5 f (3) 5 f (27)

5 8 2 5(1) 5 8 2 5(3) 5 8 2 5(27)

5 3 5 27 5 43

31. f (x) 5 x2 1 2, x0 5 21

x1 5 f (x0) x2 5 f (x1) x3 5 f (x2)

5 f (21) 5 f (3) 5 f (11)

5 (21)2 1 2 5 32 1 2 5 112 1 2

5 3 5 11 5 123

32. a. n represents the number of rows and columns.

an represents the number of blue squares.

b. n 1 2 3 4 5 6 7 8

an 1 2 5 8 13 18 25 32

c. an 5 n2

} 2 1 1 } 4 [1 2 (21)n]

a1 5 12

} 2 1 1 } 4 [1 2 (21)1] 5 1

a2 5 22

} 2 1 1 } 4 [1 2 (21)2] 5 2

a3 5 32

} 2 1 1 } 4 [1 2 (21)3] 5 5

a4 5 42

} 2 1 1 } 4 [1 2 (21)4] 5 8

a5 5 52

} 2 1 1 } 4 [1 2 (21)5] 5 13

a6 5 62

} 2 1 1 } 4 [1 2 (21)6] 5 18

a7 5 72

} 2 1 1 } 4 [1 2 (21)7] 5 25

a8 5 82

} 2 1 1 } 4 [1 2 (21)8] 5 32

These values are the same as the values in the table, so this rule defi nes the sequence represented by the checkerboard quilts.

33. a1 5 3072, r 5 1 } 4

an 5 a1rn 2 1

an 5 3072 1 1 } 4 2 n 2 1

There are a5 5 3072 1 1 } 4 2 5 2 1

5 12 actors in the

fi fth round of auditions. After this round, three actors will remain, which is exactly the number required for the three main parts in the play. So, the rule makes sense for 1 ≤ n ≤ 5.

Standardized Test Preparation (p. 845)

1. B;

an 5 2n 2 5

a15 5 2(15) 2 5 5 25

2. A;

Sn 5 n 1 a1 1 an }

2 2

S14 5 14 1 a1 1 a14 }

2 2

S14 5 14 1 23 1 23 }

2 2 5 140

Standardized Test Practice (pp. 846–847)

1. C;

S 5 a1 } 1 2 r

Basketball: a1 5 7.2, r 5 0.36

S 5 7.2 } 1 2 0.36 5 11.25

Baseball: a1 5 6, r 5 0.30

S 5 6 } 1 2 0.3 ø 8.57

Difference 5 11.25 2 8.57 5 2.68

2. C;

a1 5 20, r 5 10

} 20 5 0.5

an 5 ran 2 1

a1 5 20, an 5 0.5an 2 1

3. D;

The sequence is 12, 11, 9, 6, 2, 23 . . ..

There is no common difference or common ratio, so this sequence is neither geometric nor arithmetic.

4. B;

a1 5 1 5 20

a2 5 2 5 21

a3 5 4 5 22

a4 5 8 5 23

an 5 2n 2 1


n2ws-1200-b.indd 707 6/27/06 11:36:54 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.


5. D;

a1 5 1, r 5 2

Sn 5 a1 1 1 2 rn }

1 2 r 2

S8 5 a1 1 1 2 r8 }

1 2 r 2

S8 5 1 1 1 2 28 }

1 2 2 2 5 255

6. B;

a1 5 2 5 1 p 2

a2 5 6 5 2 p 3

a3 5 12 5 3 p 4

a4 5 20 5 4 p 5

an 5 n(n 1 1)

7. D;

n 5 4, an 5 n(n 1 1)

Sum 5 i 5 1

∑ 4

i(i 1 1)

8. a3 5 12, a5 5 48

a3 5 a1r3 2 1 → 12 5 a1r2 → a1 5 12

} r2

a5 5 a1r5 2 1 → 48 5 a1r4

48 5 1 12 }

r2 2 r4

48 5 12r2

4 5 r2

62 5 r

a1 5 12 }

(62)2 → a1 5 3

9. The sequence is geometric.

a1 5 24, r 5 12

} 24 5 23

an 5 a1rn 2 1

an 5 24(23)n 2 1

a8 5 24(23)8 2 1 5 8748

10. i 5 1

∑ 5

0.5(2)i 2 1 5 0.5 1 1 1 2 1 4 1 8 5 15.5

11. 0.151515. . . 5 15(0.01) 1 15(0.01)2 1 15(0.01)3 1 . . .

5 a1 } 1 2 r

5 15(0.01)

} 1 2 0.01

5 0.15

} 0.99

5 15

} 99

5 5 } 33

12. f (x) 5 2x 2 1, x0 5 2

x1 5 f (x0) x2 5 f (x1) x3 5 f (x2)

5 f (2) 5 f (3) 5 f (5)

5 2(2) 2 1 5 2(3) 2 1 5 2(5) 2 1

5 3 5 5 5 9

x1 1 x2 1 x3 5 3 1 5 1 9 5 17

13. a1 5 0.5 a5 5 2(39) 1 5 5 83

a2 5 2(0.5) 1 5 5 6 a6 5 2(83) 1 5 5 171

a3 5 2(6) 1 5 5 17 a7 5 2(171) 1 5 5 347

a4 5 2(17) 1 5 5 39 a8 5 2(347) 1 5 5 699

14. an 5 6n 1 3

a1 5 6(1) 1 3 5 9, a15 5 6(15) 1 3 5 93

Sn 5 n 1 a1 1 an }

2 2

S15 5 15 1 9 1 93 }

2 2 5 765

15. d 5 6 2 3 5 3

16. First round 5 10 people

Second round 5 10 p 10 people

Third round 5 10 p 10 p 10 people

nth round 5 10n people

If 100 million people already received the email, then you are in the 8th round because 108 5 100 million.

17. dn 5 1 } 2 n(n 2 3), n ≥ 3

d3 5 1 } 2 (3)(3 2 3) 5 0

d4 5 1 } 2 (4)(4 2 3) 5 2

d5 5 1 } 2 (5)(5 2 3) 5 5

d6 5 1 } 2 (6)(6 2 3) 5 9

d7 5 1 } 2 (7)(7 2 3) 5 14

d8 5 1 } 2 (8)(8 2 3) 5 20

There is no common ratio or common difference between consecutive terms, so the sequence is neither geometric nor arithmetic.

18. This situation should be represented by a series.

Sample answer: The company is pledging a total amount, not just a sequence of values.

19. a. The sequence is arithmetic because each radius is 1.22 meters greater than the previous radius.

b. a1 5 36.5, d 5 1.22

an 5 a1 1 (n 2 1)d

an 5 36.5 1 (n 2 1)(1.22) 5 35.28 1 1.22n

c. a8 5 35.28 1 1.22(8) 5 45.04

The track meets the requirement because the curve radius of lane 8 is 45.04 meters, which is less than the 50-meter maximum.


n2ws-1200-b.indd 708 6/27/06 11:36:58 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.

709Algebra 2


20. a. Current balance

5 Interest p Previous balance 2 Payment

an 5 1.0075 p an 2 1 2 300

a1 5 16,000, an 5 1.0075an 2 1 2 300

b. a18 ø 12,749.33

Mark will owe $12,749.33 at the beginning of the 18th month.

c. Mark will pay the loan off in 69 months, with the last payment being about $109.00.

d. If the payments are $350, then Mark can have the loan paid off in 57 months. He will pay less overall because the balance is less each month and the balance is paid off faster, so less interest is paid.

Cumulative Review, Chs. 1–12 (pp. 848–849)

1. 3x 2 y 5 5 2. 1 }

2 x 1 3y 5 24

1

x

y

21

1

x

y

21

3. y 5 x 1 3 2 8 4. y 5 x2 2 6x 2 27

1

x

y

21

5

x

y

24

5. y 5 22(x 1 6)(x 2 1) 6. y 5 (x 2 3)2 1 4

5

x

y

21

1

x

y

21

7. y 5 Ï}

x 1 6 8. y 5 3 Ï}

x 2 2

1

x

y

21

1

x

y

21

9. y 5 3 p 4x 2 2 10. y 5 12 1 1 } 8 2

x

1

x

y

21

1

x

y

21

11. y 5 2 } x 2 3 1 5 12. y 5

6 }

x2 2 4

1

x

y

21

x

y

21

1

13. 2 3 1 8 5 2(8) 2 1(3) 5 13

14. 12 3 27 8 5 12(8) 2 (27)(3) 5 117

15. 0 5 2 10 13 24

25 4 21 0 5

10 13

25 4

5 (0 1 100 1 80) 2 (2130 1 0 2 50)

5 360

16. 5 29 4 4 2 1

0 1 1 5 29

4 2

0 1

5 (10 1 0 1 16) 2 (0 1 5 2 36)

5 57

17. y 5 a } x

6 5 a } 3 → a 5 18

y 5 18

} x

When x 5 28: y 5 18

} 28 5 2

9 } 4

18. y 5 a } x

9 5 a }

24 → a 5 236

y 5 236

} x

When x 5 28: y 5 236

} 28 5

9 } 2

19. y 5 a } x

1 }

8 5

a } 4 → a 5

1 } 2

y 5 1 } 2 } x 5

1 } 2x

When x 5 28: y 5 1 }

2(28) 5 2

1 } 16


n2ws-1200-b.indd 709 6/27/06 11:37:12 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.


20. y 5 a } x

2 } 5 5

a } 9 → a 5

18 } 5

y 5 18

} 5 } x 5

18 } 5x

When x 5 28: y 5 18 }

5(28) 5 2

9 } 20

21. x2

} 36

1 y2

} 4 5 1 4

x

y

22

x2 }

62 1 y2

} 22 5 1

Vertices: (6, 0), (26, 0)

Co-vertices: (0, 2), (0, 22)

22. y2

} 100

2 x2

} 49 5 1

x

y

22

4

y2

} 102 2

x2

} 72 5 1

Transverse axis: Vertical

Asymptotes: y 5 10

} 7 x,

y 5 2 10

} 7 x

Vertices: (0, 10), (0, 210)

23. (x 2 3)2 5 16y

2

x

y

21

(x 2 3)2 5 16(y 1 0)

Vertex: (3, 0)

24. 9P3 5 9! }

(9 2 3)! 5

9! } 6! 5 504

25. 16P5 5 16! }

(16 2 5)! 5

16! } 11! 5 524,160

26. 7C2 5 7! } 5! p 2! 5 21

27. 6C6 5 6! } 6! p 0! 5

6! } 6! 5 1


0.85 5 0.32 1 0.6 2 P(A and B)

P(A and B) 5 0.32 1 0.6 2 0.85 5 0.07

29. P(A and B) 5 P(A) p P(BA) P(A and B) 5 0.5 p 0.3 5 0.15

30. P(A and B) 5 P(A) p P(B)

0.2 5 0.25 p P(B)

0.8 5 P(B)

31. Mean: } x 5 19 1 11 1 8 1 10 1 11 1 15 1 16

}}} 7 ø12.857

Median: 11

Mode: 11

Range: 19 2 8 5 11

Std. Dev:

� 5 Î}}}}}

(19 2 12.857)2 1 (11 2 12.857)2 1 . . . 1 (16 2 12.857)2

}}}}} 7

ø 3.603

32. Mean: } x 5 54 1 58 1 49 1 60 1 63 1 58 1 42

}}} 7 ø 54.857

Median: 58

Mode: 58

Range: 63 2 42 5 21

Std. Dev:

� 5 Î}}}}}

(54 2 54.857)2 1 (58 2 54.857)2 1 . . . 1 (42 2 54.857)2

}}}}} 7

ø 6.685

33. Mean: } x 5 216 1 203 1 225 1 216 1 212 1 228 1 209

}}}} 7

ø 215.57

Median: 216

Mode: 216

Range: 228 2 203 5 25

Std. Dev:

5 Î}}}}}

(216 2 215.57)2 1 (203 2 215.57)2 1 . . . 1 (209 2 215.57)2

}}}}} 7

ø 8.086

34. Mean: } x 5 23 1 5 2 11 1 6 2 3 1 2

}} 6 5 2 2 } 3

Median: 23 1 2

} 2 5 2

1 } 2

Mode: 23

Range: 6 2 (211) 5 17

Std. Dev:

5 Î}}}}

1 23 2 1 2

2 } 3 2 2 2 1 1 5 2 1 2

2 } 3 2 2 2 1 . . . 1 1 2 2 1 2 } 3 2 2 2

}}}} 6

ø 5.793

35. Mean: } x 5 99 1 92 1 93 1 82 1 88 1 71 1 97

}}} 7 ø 88.857

Median: 92

Mode: none

Range: 99 2 71 5 28

Std. Dev:

� 5 Î}}}}}

(99 2 88.857)2 1 (92 2 88.857)2 1 . . . 1 (97 2 88.857)2

}}}}} 7

ø 8.967


n2ws-1200-b.indd 710 6/27/06 11:37:19 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.

711Algebra 2


36. Mean: } x 5 78 1 4 1 28 1 57 1 88 1 24 1 57 1 37 1 65

}}}} 9

ø 48.667

Median: 57

Mode: 57

Range: 88 2 4 5 84

Std. Dev:

5 Î}}}}}

(78 2 48.667)2 1 (4 2 48.667)2 1 . . . 1 (65 2 48.667)2

}}}}} 9

ø 25.777

37. i 5 1

∑ 6

3i2 5 3(12) 1 3(22) 1 3(32) 1 3(42) 1 3(52) 1 3(62)

5 273

38. i 5 1

∑ 16

(22 1 i) 5 (22 1 1) 1 (22 1 2) 1 . . .

1 (22 1 15) 1 (22 1 16) 5 104

39. a1 5 1, r 5 2 } 3

Sn 5 a1 1 1 2 rn }

1 2 r 2

S12 5 1 1 1 2 1 2 } 3 2 12

} 1 2

2 } 3 2 ø 2.977

40. a1 5 5, r 5 1 } 3

S 5 a1 } 1 2 r

S 5 5 }

1 2 1 } 3 5 7.5

41. Explicit rule:

a1 5 27, d 5 23 2 (27) 5 4

an 5 a1 1 (n 2 1)d

an 5 27 1 (n 2 1)(4) 5 211 1 4n

Recursive rule:

an 5 an 2 1 1 d

a1 5 27, an 5 an 2 1 1 4

42. Explicit rule:

a1 5 1, d 5 214 2 1 5 215

an 5 a1 1 (n 2 1)d

an 5 1 1 (n 2 1)(215) 5 16 2 15n

Recursive rule:

an 5 an 2 1 1 d

a1 5 1, an 5 an 2 1 2 15

43. Explicit rule:

a1 5 3, r 5 12

} 3 5 4

an 5 a1rn 2 1

an 5 3(4)n 2 1

Recursive rule:

an 5 ran 2 1

a1 5 3, an 5 4an 2 1

44. x 5 bags of cookies

Bags of cookies

Cal

end

ars

0 300 600 900 1200 x

y

0

100

200

300 y 5 calendars

2x 1 7y 5 2400

When y 5 200:

2x 1 7(200) 5 2400

2x 5 1000

x 5 500

If you sell 200 calendars, you will need to sell 500 bags of cookies to meet your goal.

45. V 5 75π, h 5 r 1 4

V 5 πr2h

} 3

75π 5 πr 2(r 1 4)

} 3

225π 5 πr2(r 1 4)

225 5 r3 1 4r2

0 5 r3 1 4r2 2 225

Possible zeros: 61, 63, 65, 69, 615, 625, 645, 675, 6225

Test r 5 1:

1 1 4 0 2225

1 5 5

1 5 5 2220 → 1 is not a zero.

Test r 5 5:

5 1 4 0 2225

5 45 225

1 9 45 0 → 5 is a zero.

r3 1 4r2 2 225 5 (r 2 5)(r2 1 9r 1 45) Because r2 1 9r 1 45 does not have any real roots,

r 5 5 is the only zero. The radius of the cone should be 5 inches.

46. I 5 17, R 5 6.5

I 5 Î}

P

} R

17 5 Î}

P }

6.5

289 5 P } 6.5

1878.5 5 P

The hair dryer consumes 1878.5 watts of power.


n2ws-1200-b.indd 711 6/27/06 11:37:24 AM

Copy

right

© b

y M

cDou

gal L

ittel

l, a

divi

sion

of H

ough

ton

Miffl

in C

ompa

ny.


47. a1 5 18,600(0.845) 5 15,717, r 5 0.845

an 5 a1rn 2 1

an 5 15,717(0.845)n 2 1

8000 5 15,717(0.845)n 2 1

0.509 ø 0.845n 2 1

log 0.509 ø (n 2 1) log 0.845

4 ø n 2 1

5 ø n

The car will be worth $8000 in about 5 years.

48. Let the center of the circle be (0, 0) with radius 330

} 2 ,

or r 5 165.

x2 1 y2 5 r2

x2 1 y2 5 1652

x2 1 y2 5 27,225

When y 5 115:

25

25

x

y

(0, 115)

50

(0, 165)

(20 35, 115)(220 35, 115)

x2 1 1152 5 27,225

x2 5 14,000

x 5 620 Ï}

35

Distance between (220 Ï}

35 , 115) and (20 Ï}

35 , 115):

d 5 Ï}}}}

(20 Ï}

35 2 (220 Ï}

35 ))2 1 (115 2 115)2

5 Ï}

(40 Ï}

35 )2

5 40 Ï}

35

ø 236.64

The rope is about 237 feet long.

49. Let events A, B, C, and D be choosing item 1, item 2, item 3, and item 4, respectively, so that all items are different.

P(A) p P(B) p P(C) p P(D) 5 20

} 20 p 19 }

20 p 18

} 20

p 17 }

20

5 116,280

} 160,000 ø 0.727

The probability that all four people order a different item is about 0.727, or 72.7%.

50. a. Mean:

ø $202,957.14

Median: $201,900

Mode: none

Range: $23,400

Std. Dev.:

s 5 Î}}}}}

(201,900 2 202,957.14)2 1 . . . 1 (192,100 2 202,957.14)2

}}}}} 7

ø $7519.85

b. Commissions:

$201,900(0.05) 5 $10,095

$205,200(0.05) 5 $10,260

$195,800(0.05) 5 $9790

$210,300(0.05) 5 $10,515

$199,900(0.05) 5 $9995

$215,500(0.05) 5 $10,775

$192,100(0.05) 5 $9605

Mean:

x 5 10,095 1 10,260 1 9790 1 10,515 1 9995 1 10,775 1 9605

}}}}} 7

ø $10,147.86

Median: $10,095

Mode: none

Range: $1170

Std. Dev.:

s 5 Ï}}}}

(10,095 2 10,147.86)2 1 . . . 1 (9605 2 10,147.86)2

}}}} 7

ø $375.99

c. Each statistical measure in part (b) is 5% of the corresponding statistical measure in part (a).

51. a1 5 31,000, d 5 1600

an 5 a1 1 (n 2 1)d

an 5 31,000 1 (n 2 1)(1600) 5 29,400 1 1600n

a9 5 29,400 1 1600(9) 5 43,800

In the ninth year, the accountant’s salary will be $43,800.


} x 5

201,900 1 205,200 1 195,800 1 210,300 1 199,900 1 215,500 1 192,100

}}}}}} 7

n2ws-1200-b.indd 712 6/27/06 11:37:27 AM

Download - Aat Solutions - Ch12

Top Related