chapter 12 sequences; induction; the binomial theoremshubbard/hpcanswers/pr_9e_ism_12all.pdf · 33...
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Chapter 12 Sequences; Induction; the Binomial Theorem
Section 12.1
1. ( ) 2 1 12
2 2f
−= = ; ( ) 3 1 23
3 3f
−= =
2. True
3. sequence
4. True
5. ( 1) 3 2 1n n − ⋅⋅ ⋅ ⋅ ⋅
6. Recursive
7. summation
8. True; This is the sum of the first n intergers.
9. 10! 10 9 8 7 6 5 4 3 2 1 3,628,800= ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ =
10. 9! 9 8 7 6 5 4 3 2 1 362,880= ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ =
11. 9! 9 8 7 6!
9 8 7 5046! 6!
⋅ ⋅ ⋅= = ⋅ ⋅ =
12. 12! 12 11 10!
12 11 13210! 10!
⋅ ⋅= = ⋅ =
13. 3! 7! 3 2 1 7 6 5 4!
4! 4!3 2 1 7 6 5 1, 260
⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅=
= ⋅ ⋅ ⋅ ⋅ ⋅ =
14. 5! 8! 5 4 3! 8!
3! 3!5 4 8 7 6 5 4 3 2 1
806,400
⋅ ⋅ ⋅ ⋅=
= ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅=
15. 1 2 3 4 51, 2, 3, 4, 5s s s s s= = = = =
16. 2 2 21 2 3
2 24 5
1 1 2, 2 1 5, 3 1 10,
4 1 17, 5 1 26
s s s
s s
= + = = + = = + =
= + = = + =
17. 1 2
3 4
1 1 2 2 1, ,
1 2 3 2 2 4 23 3 4 4 2
, ,3 2 5 4 2 6 3
a a
a a
= = = = =+ +
= = = = =+ +
55 5
5 2 7a = =
+
18. 1 2
3 4
2 1 1 3 2 2 1 5, ,
2 1 2 2 2 42 3 1 7 2 4 1 9
, ,2 3 6 2 4 8
b b
b b
⋅ + ⋅ += = = =⋅ ⋅
⋅ + ⋅ += = = =⋅ ⋅
52 5 1 11
2 5 10b
⋅ += =⋅
19. 1 1 2 2 1 21 2
3 1 2 4 1 23 4
( 1) (1 ) 1, ( 1) (2 ) 4,
( 1) (3 ) 9, ( 1) (4 ) 16,
c c
c c
+ +
+ +
= − = = − = −
= − = = − = −
5 1 25 ( 1) (5 ) 25c += − =
20. 1 11
2 12
3 13
1( 1) 1,
2 1 1
2 2( 1) ,
2 2 1 3
3 3( 1) ,
2 3 1 5
d
d
d
−
−
−
= − = ⋅ − = − = − ⋅ − = − = ⋅ −
4 14
5 15
4 4( 1) ,
2 4 1 7
5 5( 1)
2 5 1 9
d
d
−
−
= − = − ⋅ − = − = ⋅ −
21. 1 2
1 21 2
3 4
3 43 4
5
5 5
2 2 1 2 4 2, ,
4 2 10 53 1 3 1
2 8 2 2 16 8, ,
28 7 82 413 1 3 1
2 32 8
244 613 1
s s
s s
s
= = = = = =+ +
= = = = = =+ +
= = =+
Section 12.1: Sequences
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22. 1 2
1 2
3 4
3 4
5
5
4 4 4 16, ,
3 3 3 9
4 64 4 256, ,
3 27 3 81
4 1024
3 243
s s
s s
s
= = = =
= = = =
= =
23. 1
1( 1) 1 1
,(1 1)(1 2) 2 3 6
t− −= = = −
+ + ⋅
2
2
3
3
4
4
5
5
( 1) 1 1,
(2 1)(2 2) 3 4 12
( 1) 1 1,
(3 1)(3 2) 4 5 20
( 1) 1 1,
(4 1)(4 2) 5 6 30
( 1) 1 1
(5 1)(5 2) 6 7 42
t
t
t
t
−= = =+ + ⋅
− −= = = −+ + ⋅
−= = =+ + ⋅
− −= = = −+ + ⋅
24. 1 2 3
1 2 3
4 5
4 5
3 3 3 9 3 273, , 9,
1 1 2 2 3 3
3 81 3 243,
4 4 5 5
a a a
a a
= = = = = = = =
= = = =
25. 1 2 31 2 3
4 54 5
1 1 2 3, , ,
4 5,
b b bee e e
b be e
= = = =
= =
26. 2 2 2
1 2 31 2 3
2 2
4 54 5
1 1 2 3 9, 1, ,
2 82 2 2
4 16 5 251,
16 322 2
c c c
c c
= = = = = =
= = = = =
27. Each term is a fraction with the numerator equal to the term number and the denominator equal to one more than the term number.
1nn
an
=+
28. Each term is a fraction with the numerator equal to 1 and the denominator equal to the product of the term number and one more than the term number.
( )1
1nan n
=+
29. Each term is a fraction with the numerator equal to 1 and the denominator equal to a power of 2. The power is equal to one less than the term number.
1
1
2n n
a −=
30. Each term is equal to a fraction with the numerator equal to a power of 2 and the denominator equal to a power of 3. Both powers are equal to the term number. Since the powers are the same, we can use rules for exponents to
write each term as a power of 23 .
2
3
n
na =
31. The terms form an alternating sequence. Ignoring the sign, each term always contains a 1. The sign alternates by raising 1− to a power. Since the first term is positive, we use 1n + as the power.
( ) 11
nna
+= −
32. The terms appear to alternate between whole numbers and fractions. If we write the whole
numbers as fractions (e.g. 11 1= , 33 1= , etc.), we
see that each term consists of a 1 and the term number. When n is odd, the numerator is n and the denominator is 1. When n is even, the numerator is 1 and the denominator is n. This alternating behavior occurs if we have a power that alternates sign. The alternating sign is
obtained by using ( ) 11
n−− . Thus, we get
( ) 11
n
na n−−=
33. The terms (ignoring the sign) are equal to the term number. The alternating sign is obtained by
using ( ) 11
n+− .
( ) 11
nna n
+= − ⋅
34. Here again we have alternating signs which will
be taken care of by using ( ) 11
n−− . The rest of the
term is twice the term number.
( ) 11 2
nna n
−= − ⋅
35. 1 2 3
4 5
2, 3 2 5, 3 5 8,
3 8 11, 3 11 14
a a a
a a
= = + = = + == + = = + =
Chapter 12: Sequences; Induction; the Binomial Theorem
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36. 1 2 3
4 5
3, 4 3 1, 4 1 3,
4 3 1, 4 1 3
a a a
a a
= = − = = − == − = = − =
37. 1 2 3
4 5
2, 2 ( 2) 0, 3 0 3,
4 3 7, 5 7 12
a a a
a a
= − = + − = = + == + = = + =
38. 1 2 3
4 5
1, 2 1 1, 3 1 2,
4 2 2, 5 2 3
a a a
a a
= = − = = − == − = = − =
39. 1 2 3
4 5
5, 2 5 10, 2 10 20,
2 20 40, 2 40 80
a a a
a a
= = ⋅ = = ⋅ == ⋅ = = ⋅ =
40. 1 2 3
4 5
2, 2, ( 2) 2,
2, ( 2) 2
a a a
a a
= = − = − − == − = − − =
41. 1 2 3
4 5
33 123, , ,2 3 2
111 182 ,
4 8 5 40
a a a
a a
= = = =
= = = =
42. 1 2
3 4
2, 2 3( 2) 4,
3 3( 4) 9, 4 3( 9) 23,
a a
a a
= − = + − = −= + − = − = + − = −
5 5 3( 23) 64a = + − = −
43. 1 2 3 4
5
1, 2, 2 1 2, 2 2 4,
4 2 8
a a a a
a
= = = ⋅ = = ⋅ == ⋅ =
44. 1 2 3
4 5
1, 1, 1 3 1 2,
1 4 2 9, 2 5 9 47
a a a
a a
= − = = − + ⋅ == + ⋅ = = + ⋅ =
45. 1 2 3
4
5
, , ( ) 2 ,
( 2 ) 3 ,
( 3 ) 4
a A a A d a A d d A d
a A d d A d
a A d d A d
= = + = + + = += + + = += + + = +
46.
( ) ( )2
1 2 3
2 3 3 44 5
, , ( ) ,
,
a A a rA a r rA r A
a r r A r A a r r A r A
= = = =
= = = =
47. 1 2 3
4
2, 2 2 , 2 2 2 ,
2 2 2 2 ,
a a a
a
= = + = + +
= + + +
5 2 2 2 2 2a = + + + +
48. 1 2 3
22 22, , ,
2 2a a a= = =
4 5
22 2
2 22 2,
2 2a a= =
49. ( )1( 2) 3 4 5 6 7 2
n
kk n
=+ = + + + + + ⋅⋅⋅ + +
50. ( )1(2 1) 3 5 7 9 2 1
n
kk n
=+ = + + + + ⋅⋅⋅ + +
51. 2 2
1
1 9 25 492 8 18 32
2 2 2 22 2
n
k
k n
=+ + + + + + + + ⋅⋅ ⋅ +=
52. ( ) ( )2 2
11 4 9 16 25 36 1
n
kk n
=+ = + + + + + ⋅⋅⋅ + +
53. 0
1 1 1 1 11
3 9 273 3
n
k nk =
= + + + + +
54. 0
3 3 9 31
2 2 4 2
k nn
k =
= + + + +
55. 1
10
1 1 1 1 1
3 9 273 3
n
k nk
−
+=
= + + + +
56. ( )1
0(2 1) 1 3 5 7 2( 1) 1
1 3 5 7 (2 1)
n
kk n
n
−
=+ = + + + + + − +
= + + + + + −
57. 2( 1) ln ln 2 ln 3 ln 4 ( 1) ln
nk n
kk n
=− = − + − + −
58. 1
3
4 3 5 4 6 5 1
( 1) 2
( 1) 2 ( 1) 2 ( 1) 2 ( 1) 2
nk k
k
n n
+
=+
−
= − + − + − + + −
3 4 5 6 1
1
2 2 2 2 ( 1) 2
8 16 32 64 ... ( 1) 2
n n
n n
+
+
= − + − + + −
= − + − + + −
59. 20
11 2 3 20
kk
=+ + + + =
Section 12.1: Sequences
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60. 8
3 3 3 3 3
11 2 3 8
kk
=+ + + + =
61. 13
1
1 2 3 13
2 3 4 13 1 1k
k
k=+ + + + =
+ +
62. [ ]12
11 3 5 7 2(12) 1 (2 1)
kk
=+ + + + + − = −
63. 6
66
0
1 1 1 1 11 ( 1) ( 1)
3 9 27 3 3k
kk =
− + − + + − = −
64. 11 11
11 1 1
1
2 4 8
3 9 27
2 2
3 3( 1) ( 1)
kk
k
+ +
=− + + + − = −
65. 2 3
1
3 3 3 33
2 3
n kn
kn k=+ + + + =
66. 2 3
1
1 2 3 n
n kk
n k
e e e e e=+ + + + =
67.
( )0
1
( ) ( 2 ) ( ) ( )
or ( 1 )
n
k
n
k
a a d a d a nd a kd
a k d
=
=
+ + + + + + + = +
= + −
68. 2 1 1
1
nn k
ka a r a r a r a r− −
=+ + + + =
69. ( )40
1 40 times
5 5 5 5 5 40 5 200k =
= + + + ⋅⋅⋅ + = =
70. 50
1 50 times
8 8 8 8 8 50(8) 400k =
= + + + ⋅⋅⋅+ = =
71. ( ) ( )
40
1
40 40 120 41 820
2kk
=
+= = =
72. ( )24 24
1 1
24 24 1( ) 300
2k kk k
= =
+− = − = − = −
73.
( ) ( )
20 20 20 20 20
1 1 1 1 1(5 3) (5 ) 3 5 3
20 20 15 3 20
2
1050 60 1110
k k k k kk k k
= = = = =+ = + = +
+ = +
= + =
74. ( ) ( )
( ) ( )
26 26 26 26 26
1 1 1 1 13 7 3 7 3 7
26 26 13 7 26
2
1053 182 871
k k k k kk k k
= = = = =− = − = −
+ = −
= − =
75. ( )( )( ) ( )
16 16 162 2
1 1 14 4
16 16 1 2 16 14 16
61496 64 1560
k k kk k
= = =+ = +
+ ⋅ += +
= + =
76. ( ) ( ) ( )
( ) ( ) ( )
14 142 2 2
0 1
14 142
1 1
4 0 4 4
4
14 14 1 2 14 14 4 14
64 1015 64 955
k k
k k
k k
k
= =
= =
− = − + −
= −
+ ⋅ += − + −
= − + − =
77.
( ) ( )
[ ]
60 60 60 9
10 10 1 12 2 2 2
60 60 1 9 9 12
2 2
2 1830 45 3570
k k k kk k k k
= = = =
= = −
+ + = −
= − =
78. ( )
( ) ( )
[ ]
40 40 40 7
8 8 1 13 3 3
40 40 1 7 7 13
2 2
3 820 28 2376
k k k kk k k k
= = = =
− = − = − −
+ + = − −
= − − = −
79.
( ) ( )
20 20 43 3 3
5 1 1
2 2
2 2
20 20 1 4 4 1
2 2
210 10 44,000
k k kk k k
= = == −
+ + = −
= − =
80.
( ) ( )
24 24 33 3 3
4 1 1
2 2
2 2
24 24 1 3 3 1
2 2
300 6 89,964
k k kk k k
= = == −
+ + = −
= − =
Chapter 12: Sequences; Induction; the Binomial Theorem
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81. 1 01.01 100 1.01(3000) 100 $2930B B= − = − =
John’s balance is $2930 after making the first payment.
82. 1 01.03 20 1.03(2000) 20 2080p p= + = + =
2 11.03 20 1.03(2080) 20 2162.4p p= + = + =
There are approximately 2162 trout in the pond after 2 months.
83. 1 01.005 534.47
1.005(18,500) 534.47
$18,058.03
B B= −= −=
Phil’s balance is $18,058.03 after making the first payment.
84. 1 00.9 15 0.9(250) 15 240p p= + = + =
2 10.9 15 0.9(240) 15 231p p= + = + =
There are 231 tons of pollutants after two years.
85. 1 2 3 4 5
6 7 8 1 2
1, 1, 2, 3, 5,
8, 13, 21, n n n
a a a a a
a a a a a a− −
= = = = == = = = +
8 7 6 13 8 21a a a= + = + =
After 7 months there are 21 mature pairs of rabbits.
86. a. ( ) ( )1 1
1 1
1 5 1 5 1 5 1 5 2 51
2 5 2 5 2 5u
+ − − + − += = = =
( ) ( )2 2
2 2
1 5 1 5 1 2 5 5 1 2 5 5 4 51
2 5 4 5 4 5u
+ − − + + − + −= = = =
b. ( ) ( ) ( ) ( )1 1
1 1
1 5 1 5 1 5 1 5
2 5 2 5
n n n n
n n n nu u
+ +
+ +
+ − − + − −+ = +
( ) ( ) ( ) ( )
( ) ( )
( ) ( )( ) ( )
( )( ) ( )
( )
( ) ( )( ) ( ) ( )
( ) ( ) ( )
( ) ( )
1 1
1
1
2 2
2 2
1 1
2 22 2
1 1
2 2
2
1 5 1 5 2 1 5 2 1 5
2 5
1 5 1 5 2 1 5 1 5 2
2 5
3 5 3 51 5 1 5
1 5 3 5 1 5 3 5 1 5 1 5
2 5 2 5
3 5 3 51 5 1 5 1 1
1 5 1 56 2 5 6 2 52 2
2 5 2 5
1 5 1 5
2
n n n n
n
n n
n
n n
n n
n n
n nn n
n n
n n
n
+ +
+
+
+ +
+ +
+ ++ +
+ +
+ +
+
+ − − + + − −=
+ + + − − − + =
+ −+ − −
+ + − − − + − = =
+ −+ − −
+ − −+ −= =
+ − −=
2
5
nu +=
c. Since { }1 2 2 11, 1, ,n n n nu u u u u u+ += = = + is the Fibonacci sequence.
Section 12.1: Sequences
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87. 1, 1, 2, 3, 5, 8, 13 This is the Fibonacci sequence.
88. a. 1 2 3 4 5 6 7 81, 1, 2, 3, 5, 8, 13, 21,u u u u u u u u= = = = = = = = 9 1034, 55u u= = , 11 89u =
b. 3 5 62 4
1 2 3 4 5
1 2 3 5 81, 2, 1.5, 1.67, 1.6,
1 1 2 3 5
u u uu u
u u u u u= = = = = = = ≈ = =
7 8 9
6 7 8
10 11
9 10
13 21 341.625, 1.615, 1.619,
8 13 21
55 891.618, 1.618
34 55
u u u
u u u
u u
u u
= = = ≈ = ≈
= ≈ = ≈
c. 1.618 (the exact value is 1 5
2+
)
d. 31 2 4
2 3 4 5
1 1 2 31, 0.5, 0.667, 0.6,
1 2 3 5
uu u u
u u u u= = = = = ≈ = =
5 6 7
6 7 8
8 9 10
9 10 11
5 8 130.625, 0.615, 0.619,
8 13 21
21 34 55 0.618, 0.618, 0.618
34 55 89
u u u
u u u
u u u
u u u
= = = ≈ = ≈
= ≈ = ≈ = ≈
e. 0.618 (the exact value is 2
1 5+)
89. a. ( )0 1 44
1.3 1.3!
0
1.3 1.3 1.3
0! 1! 4!1.3 ... 3.630170833
k
kk
f e=
+ + += ≈ = ≈
b. ( )0 1 77
1.3 1.3!
0
1.3 1.3 1.31.3 ... 3.669060828
0! 1! 7!
k
kk
f e=
= ≈ = + + + ≈
c. ( ) 1.31.3 3.669296668f e= ≈
d. It will take 12n = to approximate ( ) 1.31.3f e= correct to 8 decimal places.
Chapter 12: Sequences; Induction; the Binomial Theorem
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90. a. ( ) ( ) ( ) ( ) ( ) ( )0 1 2 3
2.43 2.4 2.4 2.4 2.4 2.4
0! 1! 2! 3!!0
2.4 0.824k
kk
f e− − − − − −
=− = ≈ = = −+ + +
b. ( ) ( ) ( ) ( ) ( )0 1 62.4
6 2.4 2.4 2.4 2.40! 1! 6!!
02.4 0.1602688...
k
kk
f e− − − − −
=− = ≈ == + + +
c. ( ) 2.42.4 0.0907179533f e−− = ≈
d. It will take 17n = to approximate ( ) 2.42.4f e−− = correct to 8 decimal places.
91. a. 1 0.4a = , 2 22 0.4 0.3 2 0.4 0.3 0.7a −= + ⋅ = + = , ( )3 2
3 0.4 0.3 2 0.4 0.3 2 1.0a −= + ⋅ = + = ,
( )4 24 0.4 0.3 2 0.4 0.3 4 1.6a −= + ⋅ = + = , ( )5 2
5 0.4 0.3 2 0.4 0.3 8 2.8a −= + ⋅ = + = ,
( )6 26 0.4 0.3 2 0.4 0.3 16 5.2a −= + ⋅ = + = , ( )7 2
7 0.4 0.3 2 0.4 0.3 32 10.0a −= + ⋅ = + = ,
( )8 28 0.4 0.3 2 0.4 0.3 64 19.6a −= + ⋅ = + =
The first eight terms of the sequence are 0.4, 0.7, 1.0, 1.6, 2.8, 5.2, 10.0, and 19.6.
b. Except for term 5, which has no match, Bode’s formula provides excellent approximations for the mean distances of the planets from the sun.
c. The mean distance of Ceres from the Sun is approximated by 5 2.8a = , and that of Uranus is 8 19.6a = .
d. ( )9 29 0.4 0.3 2 0.4 0.3 128 38.8a −= + ⋅ = + =
( )10 210 0.4 0.3 2 0.4 0.3 256 77.2a −= + ⋅ = + =
e. Pluto’s distance is approximated by 9a , but no term approximates Neptune’s mean distance from the sun.
f. ( )11 211 0.4 0.3 2 0.4 0.3 512 154a −= + ⋅ = + =
According to Bode’s Law, the mean orbital distance of Eris will be 154 AU from the sun.
Section 12.1: Sequences
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92. To show that ( ) ( )11 2 3 ... 1
2
n nn n
++ + + + − + =
Let ( )( ) ( )
[ ] ( ) ( ) ( ) [ ]
1 2 3 ... 1 , we can reverse the order to get
1 2 +...+ 2 + 1, now add these two lines to get
2 1 2 1 3 2 ...... 1 2 1
S n n
S n n n
S n n n n n
= + + + + − +
+ = + − + −
= + + + − + + − + + − + + +
So we have [ ] [ ] [ ] [ ] [ ] [ ]2 1 1 1 .... 1 1 1S n n n n n n n= + + + + + + + + + + = ⋅ +
( )( )
2 1
1
2
S n n
n nS
∴ = +
⋅ +=
93. 5 We begin with an initial guess of 0 2a = .
1
2
3
1 52 2.25
2 21 5
2.25 2.2361111112 2.251 5
2.2361111112 2.2361111112.236067978
a
a
a
= + = = + ≈ = +
≈
4
5
1 52.236067978
2 2.2360679782.236067977
1 52.236067977
2 2.2360679772.236067977
a
a
= +
≈ = +
≈
For both 5a and the calculator approximation,
we obtain 5 2.236067977≈ .
94. 8 We begin with an initial guess of 0 3a = .
1 00
2 11
3 22
4 33
1 8 1 83 2.833333333
2 2 3
1 8
2
1 82.833333333
2 2.2.8333333332.828431373
1 8
2
1 82.828431373
2 2.8284313732.828427125
1 8
2
12.828427125
2
a aa
a aa
a aa
a aa
= + = + ≈
= +
= +
≈
= + = +
≈
= +
= +
5 44
8
2.8284271252.828427125
1 8
2
1 82.828427125
2 2.8284271252.828427125
a aa
≈
= + = +
≈
For both 5a and the calculator approximation,
we obtain 8 2.828427125≈ .
Chapter 12: Sequences; Induction; the Binomial Theorem
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95. 21 We begin with an initial guess of 0 5a = .
1
2
3
1 215 4.625
2 51 21
4.625 4.582770272 4.6251 21
4.582770272 4.582770274.582575699
a
a
a
= + = = + ≈ = +
≈
4
1 214.582575699
2 4.5825756994.582575695
a = +
≈
5
1 214.582575695
2 4.5825756954.582575695
a = +
≈
For both 5a and the calculator approximation,
we obtain 21 4.582575695≈ .
96. 89 We begin with an initial guess of 0 9a = .
1
2
3
4
5
1 895 9.444444444
2 51 89
9.4444444442 9.4444444449.433986928
1 899.433986928
2 9.4339869289.433981132
1 899.433981132
2 9.4339811329.433981132
1 899.433981132
2 9.4339811
a
a
a
a
a
= + ≈ = +
≈ = +
≈ = +
≈
= +32
9.433981132
≈
For both 5a and the calculator approximation,
we obtain 89 9.433981132≈ .
97. 1 1u = and 1 ( 1)n nu u n+ = + + : So
1
2 1
3 2
4 3
5 4
6 5
7 6
1
(1 1) 1 2 3
(2 1) 3 3 6
(3 1) 6 4 10
(4 1) 10 5 15
(5 1) 15 6 21
(6 1) 21 7 28
u
u u
u u
u u
u u
u u
u u
== + + = + == + + = + == + + = + == + + = + == + + = + == + + = + =
98. Note that: 1 1 2 3 ( 2) ( 1) ( 1)nu n n n n+ = + + + + − + − + + +
and rewriting 1 ( 1) ( 1) ( 2) 3 2 1nu n n n n+ = + + + − + − + + + + .
So adding these together we have
1
1
1
1 2 3 ( 1) ( 1)
( 1) ( 1) 3 2 1
2( ) ( 2) ( 2) ( 2) ( 2) ( 2) ( 2)
n
n
n
u n n n
u n n n
u n n n n n n
+
+
+
= + + + + − + + += + + + − + + + +
= + + + + + + + + + + + +
1
1
2( ) ( 1)( 2)
( 1)( 2)
2
n
n
u n n
n nu
+
+
= + ++ +=
99. We know from number 97 and 98 that:
1( 1)( 2) ( )( 1)
and 2 2n n
n n n nu u+
+ + += =
Adding these together we get
1
2 2
2
22 2
( 1)( 2) ( )( 1)+
2 2( 1)( 2) ( )( 1)
2
3 2
2
2 4 2
2
2( 2 1)2 1 ( 1)
2
n nn n n n
u u
n n n n
n n n n
n n
n nn n n
++ + ++ =
+ + + +=
+ + + +=
+ +=
+ += = + + = +
100 - 101. Answers will vary.
Section 12.2: Arithmetic Sequences
1259
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Section 12.2
1. arithmetic
2. False; the sum of the first and last terms equals twice the sum of all the terms divided by the number of terms.
3. 1 1
6
12 (5 1)5 8
So
8 (6 1)5 17
a a
a
= + − = −
= − + − =
4. True
5. 1
( 4) ( 1 4) ( 4) ( 3)
4 3 1
n nd s s
n n n n
n n
−= −= + − − + = + − += + − − =
The difference between consecutive terms is constant, therefore the sequence is arithmetic.
1 2 3
4
1 4 5, 2 4 6, 3 4 7,
4 4 8
s s s
s
= + = = + = = + == + =
6.
( ) ( )1
( 5) ( 1 5) 5 6
5 6 1
n nd s s
n n n n
n n
−= −= − − − − = − − −= − − + =
The difference between consecutive terms is constant, therefore the sequence is arithmetic.
1 2 3
4
1 5 4, 2 5 3, 3 5 2,
4 5 1
s s s
s
= − = − = − = − = − = −= − = −
7.
( )( ) ( )
1
2 5 (2( 1) 5)
2 5 2 2 5
2 5 2 7 2
n nd a a
n n
n n
n n
−= −= − − − −
= − − − −= − − + =
The difference between consecutive terms is constant, therefore the sequence is arithmetic.
1 2
3 4
2 1 5 3, 2 2 5 1,
2 3 5 1, 2 4 5 3
a a
a a
= ⋅ − = − = ⋅ − = −= ⋅ − = = ⋅ − =
8.
( )( ) ( )
1
3 1 (3( 1) 1)
3 1 3 3 1
3 1 3 2 3
n nd b b
n n
n n
n n
−= −= + − − +
= + − − += + − + =
The difference between consecutive terms is constant, therefore the sequence is arithmetic.
1 2
3 4
3 1 1 4, 3 2 1 7,
3 3 1 10, 3 4 1 13
b b
b b
= ⋅ + = = ⋅ + == ⋅ + = = ⋅ + =
9.
( )( ) ( )
1
6 2 (6 2( 1))
6 2 6 2 2
6 2 6 2 2 2
n nd c c
n n
n n
n n
−= −= − − − −
= − − − += − − + − = −
The difference between consecutive terms is constant, therefore the sequence is arithmetic.
1 2
3 4
6 2 1 4, 6 2 2 2,
6 2 3 0, 6 2 4 2
c c
c c
= − ⋅ = = − ⋅ == − ⋅ = = − ⋅ = −
10.
( )( ) ( )
1
4 2 (4 2( 1))
4 2 4 2 2
4 2 4 2 2 2
n nd d d
n n
n n
n n
−= −= − − − −
= − − − += − − + − = −
The difference between consecutive terms is constant, therefore the sequence is arithmetic.
1 2
3 4
4 2 1 2, 4 2 2 0,
4 2 3 2, 4 2 4 4
d d
d d
= − ⋅ = = − ⋅ == − ⋅ = − = − ⋅ = −
11. 1
1 1 1 1( 1)
2 3 2 3
1 1 1 1 1
2 3 2 3 3
1 1 1 1 1 1
2 3 2 3 3 3
n nd t t
n n
n n
n n
−= −
= − − − − = − − − +
= − − + − = −
The difference between consecutive terms is constant, therefore the sequence is arithmetic.
1 2
3 4
1 1 1 1 1 11 , 2 ,
2 3 6 2 3 61 1 1 1 1 5
3 , 42 3 2 2 3 6
t t
t t
= − ⋅ = = − ⋅ = −
= − ⋅ = − = − ⋅ = −
12. 1
2 1 2 1( 1)
3 4 3 4
2 1 2 1 1
3 4 3 4 4
2 1 2 1 1 1
3 4 3 4 4 4
n nd t t
n n
n n
n n
−= −
= + − + − = + − + −
= + − − + =
The difference between consecutive terms is constant, therefore the sequence is arithmetic.
1 2
3 4
2 1 11 2 1 71 , 2 ,
3 4 12 3 4 62 1 17 2 1 5
3 , 43 4 12 3 4 3
t t
t t
= + ⋅ = = + ⋅ =
= + ⋅ = = + ⋅ =
Chapter 12: Sequences; Induction; the Binomial Theorem
1260
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
13.
( ) ( )( ) ( ) ( )
( ) ( ) ( )( )
1
1ln 3 ln 3
ln 3 1 ln 3
ln 3 ( 1 ) ln 3 1
ln 3
n n
n n
d s s
n n
n n n n
−−
= −
= −
= − −
= − − = − +=
The difference between consecutive terms is constant, therefore the sequence is arithmetic.
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
1 21 2
3 43 4
ln 3 ln 3 , ln 3 2ln 3 ,
ln 3 3ln 3 , ln 3 4ln 3
s s
s s
= = = =
= = = =
14. ( )ln ln( 1)1 1 1n n
n nd s s e e n n−−= − = − = − − =
The difference between consecutive terms is constant, therefore the sequence is arithmetic.
ln1 ln 2 ln31 2 3
ln 44
1, 2, 3,
4
s e s e s e
s e
= = = = = =
= =
15. 1 ( 1)
2 ( 1)3
2 3 3
3 1
na a n d
n
n
n
= + −= + −= + −= −
51 3 51 1 152a = ⋅ − =
16. 1 ( 1)
2 ( 1)4
2 4 4
4 6
na a n d
n
n
n
= + −= − + −= − + −= −
51 4 51 6 198a = ⋅ − =
17. 1 ( 1)
5 ( 1)( 3)
5 3 3
8 3
na a n d
n
n
n
= + −= + − −= − += −
51 8 3 51 145a = − ⋅ = −
18. 1 ( 1)
6 ( 1)( 2)
6 2 2
8 2
na a n d
n
n
n
= + −= + − −= − += −
51 8 2 51 94a = − ⋅ = −
19.
( )
1 ( 1)
10 ( 1)
21 1
2 21
12
na a n d
n
n
n
= + −
= + −
= −
= −
( )511
51 1 252
a = − =
20. 1 ( 1)
11 ( 1)
3
1 11
3 34 1
3 3
na a n d
n
n
n
= + −
= + − −
= − +
= −
514 1 4 51 47
513 3 3 3 3
a = − ⋅ = − = −
21. 1 ( 1)
2 ( 1) 2
2 2 2 2
na a n d
n
n n
= + −
= + −
= + − =
51 51 2a =
22. ( )1 ( 1) 0 ( 1) 1na a n d n n= + − = + − π = − π
51 51 50a = π − π = π
23. 1 12, 2, ( 1)na d a a n d= = = + −
100 2 (100 1)2 2 99(2) 2 198 200a = + − = + = + =
24. 1 11, 2, ( 1)na d a a n d= − = = + −
80 1 (80 1)2 1 79(2) 1 158 157a = − + − = − + = − + =
25. 1 11, 2 1 3, ( 1)na d a a n d= = − − = − = + −
90 1 (90 1)( 3) 1 89( 3)
1 267 266
a = + − − = + −= − = −
26. 1 15, 0 5 5, ( 1)na d a a n d= = − = − = + −
80 5 (80 1)( 5) 5 79( 5)
5 395 390
a = + − − = + −= − = −
27. 1 15 1
2, 2 , ( 1)2 2 na d a a n d= = − = = + −
801 83
2 (80 1)2 2
a = + − =
Section 12.2: Arithmetic Sequences
1261
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
28. 1
1
2 5, 4 5 2 5 2 5,
( 1)n
a d
a a n d
= = − == + −
( )70 2 5 (70 1)2 5
2 5 69 2 5
2 5 138 5
140 5
a = + −
= +
= +
=
29. 8 1 20 17 8 19 44a a d a a d= + = = + = Solve the system of equations by subtracting the first equation from the second:
1
12 36 3
8 7(3) 8 21 13
d d
a
= == − = − = −
Recursive formula: 1 113 3n na a a −= − = +
nth term: ( )( )( )
1 1
13 1 3
13 3 3
3 16
na a n d
n
n
n
= + −
= − + −= − + −= −
30. 4 1 20 13 3 19 35a a d a a d= + = = + = Solve the system of equations by subtracting the first equation from the second:
1
16 32 2
3 3(2) 3 6 3
d d
a
= == − = − = −
Recursive formula: 1 13 2n na a a −= − = +
nth term: ( )( )( )
1 1
3 1 2
3 2 2
2 5
na a n d
n
n
n
= + −
= − + −= − + −= −
31. 9 1 15 18 5 14 31a a d a a d= + = − = + = Solve the system of equations by subtracting the first equation from the second:
1
6 36 6
5 8(6) 5 48 53
d d
a
= == − − = − − = −
Recursive formula: 1 153 6n na a a −= − = +
nth term: ( )( )( )
1 1
53 1 6
53 6 6
6 59
na a n d
n
n
n
= + −
= − + −= − + −= −
32. 8 1 18 17 4 17 96a a d a a d= + = = + = − Solve the system of equations by subtracting the first equation from the second:
1
10 100 10
4 7( 10) 4 70 74
d d
a
= − = −= − − = + =
Recursive formula: 1 174 10n na a a −= = −
nth term: ( )( )( )
1 1
74 1 10
74 10 10
84 10
na a n d
n
n
n
= + −
= + − −= − += −
33. 15 1 40 114 0 39 50a a d a a d= + = = + = − Solve the system of equations by subtracting the first equation from the second:
1
25 50 2
14( 2) 28
d d
a
= − = −= − − =
Recursive formula: 1 128 2n na a a −= = −
nth term: ( )( )( )
1 1
28 1 2
28 2 2
30 2
na a n d
n
n
n
= + −
= + − −= − += −
34. 5 1 13 14 2 12 30a a d a a d= + = − = + = Solve the system of equations by subtracting the first equation from the second:
1
8 32 4
2 4(4) 18
d d
a
= == − − = −
Recursive formula: 1 118 4n na a a −= − = +
nth term: ( )( )( )
1 1
18 1 4
18 4 4
4 22
na a n d
n
n
n
= + −
= − + −= − + −= −
35. 14 1 18 113 1 17 9a a d a a d= + = − = + = − Solve the system of equations by subtracting the first equation from the second:
1
4 8 2
1 13( 2) 1 26 25
d d
a
= − = −= − − − = − + =
Recursive formula: 1 125 2n na a a −= = −
nth term: ( )( )( )
1 1
25 1 2
25 2 2
27 2
na a n d
n
n
n
= + −
= + − −= − += −
Chapter 12: Sequences; Induction; the Binomial Theorem
1262
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
36. 12 1 18 111 4 17 28a a d a a d= + = = + = Solve the system of equations by subtracting the first equation from the second:
1
6 24 4
4 11(4) 4 44 40
d d
a
= == − = − = −
Recursive formula: 1 140 4n na a a −= − = +
nth term: ( )( )( )
1 1
40 1 4
40 4 4
4 44
na a n d
n
n
n
= + −
= − + −= − + −= −
37. ( ) ( )( ) ( ) 21 1 2 1 2
2 2 2n nn n n
S a a n n n= + = + − = =
38. ( ) ( ) ( )21 2 2 1
2 2n nn n
S a a n n n n n= + = + = + = +
39. ( ) ( )( ) ( )1 7 2 5 9 52 2 2n nn n n
S a a n n= + = + + = +
40. ( ) ( )( )
( )
( )
1
2
1 4 52 2
4 6 2 32
2 3
n nn n
S a a n
nn n n
n n
= + = − + −
= − = −
= −
41. 1 12, 4 2 2, ( 1)na d a a n d= = − = = + −
70 2 ( 1)2
70 2 2 2
70 2
35
n
n
n
n
= + −= + −==
( ) ( )
( ) ( )
135
2 702 235
72 35 362
1260
n nn
S a a= + = +
= =
=
42. 1 11, 3 1 2, ( 1)na d a a n d= = − = = + −
59 1 ( 1)2
59 1 2 2
60 2
30
n
n
n
n
= + −= + −==
( ) ( ) ( )130
1 59 15 60 9002 2n nn
S a a= + = + = =
43. 1 15, 9 5 4, ( 1)na d a a n d= = − = = + −
( )49 5 1 4
49 5 4 4
48 4
12
n
n
n
n
= + −= + −==
( ) ( ) ( )112
5 49 6 54 3242 2n nn
S a a= + = + = =
44. 1 12, 5 2 3, ( 1)na d a a n d= = − = = + −
( )41 2 1 3
41 2 3 3
42 3
14
n
n
n
n
= + −= + −==
( ) ( ) ( )114
2 41 7 43 3012 2n nn
S a a= + = + = =
45. 1 73a = , 78 73 5d = − = , ( )1 1na a n d= + −
( )( )( )
558 73 1 5
485 5 1
97 1
98
n
n
n
n
= + −
= −
= −=
( ) ( )( )
198
73 5582 2
49 631 30,919
n n
nS a a= + = +
= =
46. 1 7a = , 1 7 6d = − = − , ( )1 1na a n d= + −
( )( )( )
299 7 1 6
306 6 1
51 1
52
n
n
n
n
− = + − −
− = − −
= −=
( ) ( )( )( )
152
7 2992 2
26 292 7592
n n
nS a a= + = + −
= − = −
Section 12.2: Arithmetic Sequences
1263
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
47. 1 4a = , 4.5 4 0.5d = − = , ( )1 1na a n d= + −
( )( )( )
100 4 1 0.5
96 0.5 1
192 1
193
n
n
n
n
= + −
= −
= −=
( ) ( )
( )
1193
4 1002 2193
104 10,0362
n nn
S a a= + = +
= =
48. 1 8a = , 1 1
8 84 4
d = − = , ( )1 1na a n d= + −
( )
( )
150 8 1
4
142 1
4168 1
169
n
n
n
n
= + −
= −
= −=
( ) ( )
( )
1169
8 502 2169
58 49012
n nn
S a a= + = +
= =
49. ( )1 2 1 5 3a = − = − , ( )80 2 80 5 155a = − =
( ) ( )8080
3 155 40 152 60802
S = − + = =
50. ( )1 3 2 1 1a = − = , ( )90 3 2 90 177a = − = −
( )( ) ( )9090
1 177 45 176 79202
S = + − = − = −
51. ( )11 11
6 12 2
a = − = , ( )1001
6 100 442
a = − = −
( )100100 11
442 2
7750 1925
2
S = + −
= − = −
52. ( )11 1 5
13 2 6
a = + = , ( )801 1 163
803 2 6
a = + =
( )8080 5 163
40 28 11202 6 6
S = + = =
53. 1 14a = , 16 14 2d = − = , ( )1 1na a n d= + −
( ) ( ) ( )120 14 120 1 2 14 119 2 252a = + − = + =
( ) ( )120120
14 252 60 266 15,9602
S = + = =
54. 1 2a = , 1 2 3d = − − = − , ( )1 1na a n d= + −
( ) ( ) ( )( )46 2 46 1 3 2 45 3 133a = + − − = + − = −
( )( ) ( )4646
2 133 23 131 30132
S = + − = − = −
55. Find the common difference of the terms and solve the system of equations:
(2 1) ( 3) 2
(5 2) (2 1) 3 1
3 1 2
2 3
3
2
x x d x d
x x d x d
x x
x
x
+ − + = − =+ − + = + =
+ = −= −
= −
56. Find the common difference of the terms and solve the system of equations:
(3 2) (2 ) 2
(5 3) (3 2) 2 1
2 1 2
1
x x d x d
x x d x d
x x
x
+ − = + =+ − + = + =
+ = +=
57. 13, 11, and 1092d a S= = =
[ ]
[ ][ ]
2
2
1092 2(11) ( 1)(3)2
1092 22 3 32
2194 19 3
2194 19 3
3 19 2184 0
(3 91)( 24) 0
nn
nn
n n
n n
n n
n n
= + −
= + −
= +
= +
+ − =+ − =
So 24n = .
Chapter 12: Sequences; Induction; the Binomial Theorem
1264
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
58. 14, 78, and 702d a S= − = =
[ ]
[ ][ ]
2
2
2
702 2(78) ( 1)( 4)2
702 156 4 42
1404 160 4
1404 160 4
4 160 1404 0
40 351 0
( 13)( 27) 0
nn
nn
n n
n n
n n
n n
n n
= + − −
= − +
= −
= −
− + =
− + =− − =
So 13n = or 27n = .
59. The total number of seats is:
( )( )25 26 27 25 29 1S = + + + + +
This is the sum of an arithmetic sequence with
11, 25, and 30d a n= = = .
Find the sum of the sequence:
[ ]3030
2(25) (30 1)(1)2
15(50 29) 15(79)
1185
S = + −
= + ==
There are 1185 seats in the theater.
60. The total number of seats is:
( )( )( )15 17 19 15 39 2S = + + + + +
This is the sum of an arithmetic sequence with
12, 15, and 40d a n= = = .
Find the sum of the sequence:
[ ]4040
2(15) (40 1)(2)2
20(30 78) 20(108)
2160
S = + −
= + ==
The corner section has 2160 seats.
61. The lighter colored tiles have 20 tiles in the bottom row and 1 tile in the top row. The number decreases by 1 as we move up the triangle. This is an arithmetic sequence with
1 20, 1, and 20a d n= = − = . Find the sum:
[ ]202(20) (20 1)( 1)
210(40 19) 10(21)
210
S = + − −
= − ==
There are 210 lighter tiles.
The darker colored tiles have 19 tiles in the bottom row and 1 tile in the top row. The
number decreases by 1 as we move up the triangle. This is an arithmetic sequence with
1 19, 1, and 19a d n= = − = . Find the sum:
[ ]192(19) (19 1)( 1)
219 19
(38 18) (20) 1902 2
S = + − −
= − = =
There are 190 darker tiles.
62. The number of bricks required decreases by 2 on each successive step. This is an arithmetic sequence with 1 100, 2, and 30a d n= = − = .
a. The number of bricks for the top step is:
30 1 ( 1) 100 (30 1)( 2)
100 29( 2) 100 58
42
a a n d= + − = + − −= + − = −=
42 bricks are required for the top step.
b. The total number of bricks required is the sum of the sequence:
[ ]30100 42 15(142) 2130
2S = + = =
2130 bricks are required to build the staircase.
63. The air cools at the rate of 5.5° F per 1000 feet. Since n represents thousands of feet, we have
5.5d = − . The ground temperature is 67 F° so we have 1 67 5.5 61.5T = − = . Therefore,
{ } ( )( ){ }{ } { }61.5 1 5.5
5.5 67 or 67 5.5
nT n
n n
= + − −
= − + −
After the parcel of air has risen 5000 feet, we
have ( )( )5 61.5 5 1 5.5 39.5T = + − − = .
The parcel of air will be 39.5 F° after it has risen 5000 feet.
64. If we treat the length of each rung as the term of an arithmetic sequence, we have 1 49a = ,
2.5d = − , and 24na = .
( )1 1na a n d= + −
( )( )( )
24 49 1 2.5
25 2.5 1
10 1
11
n
n
n
n
= + − −
− = − −
= −=
Therefore, the ladder contains 11 rungs.
Section 12.3: Geometric Sequences; Geometric Series
1265
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
To find the total material required for the rungs, we need the sum of their lengths. Since there are 11 rungs, we have
( ) ( )1111 11
49 24 73 401.52 2
S = + = =
It would require 401.5 feet of material to construct the rungs for the ladder.
65. 1 35a = , 37 35 2d = − = , ( )1 1na a n d= + −
( ) ( ) ( )27 35 27 1 2 35 26 2 87a = + − = + =
( ) ( )2727 27
35 87 122 16472 2
S = + = =
The amphitheater has 1647 seats.
66. Find n in an arithmetic sequence with
1 10, 4, 2040na d S= = = .
[ ]
[ ][ ]
1
2
2
2 ( 1)2
2040 2(10) ( 1)42
4080 20 4 4
4080 (4 16)
4080 4 16
1020 4
nn
S a n d
nn
n n
n n
n n
n n
= + −
= + −
= + −= +
= +
= +
2 4 1020 0
( 34)( 30) 0 34 or 30
n n
n n n n
+ − =+ − = = − =
There are 30 rows in the corner section of the stadium.
67. The yearly salaries form an arithmetic sequence with 1 35,000, 1400, 280,000na d S= = = .
Find the number of years for the aggregate salary to equal $280,000.
[ ]
[ ][ ]
1
2
2
2 ( 1)2
280,000 2(35,000) ( 1)14002
280,000 35,000 700 700
280,000 (700 34,300)
280,000 700 34,300
400 49
nn
S a n d
nn
n n
n n
n n
n n
= + −
= + −
= + −= +
= +
= +
2 49 400 0n n+ − =
249 49 4(1)( 400)
2(1)
49 4001 49 63.25
2 27.13 or 56.13
n
n n
− ± − −=
− ± − ±= ≈
≈ ≈ −
It takes about 8 years to have an aggregate salary of at least $280,000. The aggregate salary after 8 years will be $319,200.
68. Answers will vary.
69. Answers will vary. Both increase (or decrease) at a constant rate, but the domain of an arithmetic sequence is the set of natural numbers while the domain of a linear function is the set of all real numbers.
Section 12.3
1. 2 2
0.041000 1 $1082.43
2A
⋅ = + =
2. 12 1
0.0510,000 1 $9513.28
12P
− ⋅ = + =
3. geometric
4. 1
a
r−
5. divergent series
6. True
7. False; the common ratio can be positive or negative (or 0, but this results in a sequence of only 0s).
8. True
9. 1
133 3
3
nn n
nr
++ −= = =
The ratio of consecutive terms is constant, therefore the sequence is geometric.
1 21 2
3 43 4
3 3, 3 9,
3 27, 3 81
s s
s s
= = = =
= = = =
Chapter 12: Sequences; Induction; the Binomial Theorem
1266
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
10. 1
1( 5)( 5) 5
( 5)
nn n
nr
++ −−= = − = −
−
The ratio of consecutive terms is constant, therefore the sequence is geometric.
1 21 2
3 43 4
( 5) 5, ( 5) 25,
( 5) 125, ( 5) 625
s s
s s
= − = − = − =
= − = − = − =
11.
1
11
31 122 21
32
n
n n
nr
+
+ − − = = =
−
The ratio of consecutive terms is constant, therefore the sequence is geometric.
1 2
1 2
3 4
3 4
1 3 1 33 , 3 ,
2 2 2 4
1 3 1 33 , 3
2 8 2 16
a a
a a
= − = − = − = −
= − = − = − = −
12.
1
15
5 522 25
2
n
n n
nr
+
+ − = = =
The ratio of consecutive terms is constant, therefore the sequence is geometric.
1 2
1 2
3 4
3 4
5 5 5 25, ,
2 2 2 4
5 125 5 625,
2 8 2 16
b b
b b
= = = =
= = = =
13.
1 1
( 1)11
24 2
2 222
4
n
nn n
nnr
+ −
− −−−
= = = =
The ratio of consecutive terms is constant, therefore the sequence is geometric.
1 1 02
1 2
2 1 11
2 2
3 1 2
3 2
4 1 3
4 2
2 2 12 ,
4 42
2 2 12 ,
4 22
2 21,
4 2
2 22
4 2
c
c
c
c
−−
−−
−
−
= = = =
= = = =
= = =
= = =
14.
1
11
39 3
3 333
9
n
nn n
nnr
+
++ −
= = = =
The ratio of consecutive terms is constant, therefore the sequence is geometric.
1 2
1 2
3 4
3 4
3 1 3 9, 1,
9 3 9 9
3 27 3 813, 9
9 9 9 9
d d
d d
= = = = =
= = = = = =
15.
113
1/ 33 3
3
22 2
2
nn n
nr
+ + −
= = =
The ratio of consecutive terms is constant, therefore the sequence is geometric.
1/ 3 2/ 3 3/ 3 4 /31 2 3 42 , 2 , 2 2, 2e e e e= = = = =
16. 2( 1)
2 2 2 22
33 3 9
3
nn n
nr
++ −= = = =
The ratio of consecutive terms is constant, therefore the sequence is geometric.
2 1 2 2 41 2
2 3 6 2 4 83 4
3 9, 3 3 81,
3 3 729, 3 3 6561
f f
f f
⋅ ⋅
⋅ ⋅
= = = = =
= = = = = =
17.
1 1
1
1 11
( 1) ( 1) 1
3
2 3 2
3 23
2
33 2 3 2
2
n
n n n
n nn
n
n n n n
r
+ −
+
− +−
− − − + −
= = ⋅
= ⋅ = ⋅ =
The ratio of consecutive terms is constant, therefore the sequence is geometric.
1 1 0 2 1 1
1 21 2 2
3 1 2 4 1 3
3 43 3 4 4
3 3 1 3 3 3, ,
2 2 42 2 2
3 3 9 3 3 27,
8 162 2 2 2
t t
t t
− −
− −
= = = = = =
= = = = = =
Section 12.3: Geometric Sequences; Geometric Series
1267
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
18.
1
1 1 1 1
1
1 1 1
2
3 3 2
3 22
3
23 2 3 2
3
n
n n n
n nn
n
n n n n
r
+
+ − − +
−
− − + − −
= = ⋅
= ⋅ = ⋅ =
The ratio of consecutive terms is constant, therefore the sequence is geometric.
1 2
1 21 1 0 2 1
3 4
3 43 1 2 4 1 3
2 2 2 2 42, ,
1 33 3 3
2 8 8 2 16 16,
9 273 3 3 3
u u
u u
− −
− −
= = = = = =
= = = = = =
19. 5 1 45
1
2 3 2 3 2 81 162
2 3nn
a
a
−
−
= ⋅ = ⋅ = ⋅ =
= ⋅
20. 5 1 45
1
2 4 2 4 2 256 512
2 4nn
a
a
−
−
= − ⋅ = − ⋅ = − ⋅ = −
= − ⋅
21. 5 1 45
1
5( 1) 5( 1) 5 1 5
5 ( 1)nn
a
a
−
−
= − = − = ⋅ =
= ⋅ −
22. 5 1 45
1
6( 2) 6( 2) 6 16 96
6 ( 2)nn
a
a
−
−
= − = − = ⋅ =
= ⋅ −
23. 5 1 4
5
1
1 10 0 0
2 2
10 0
2
n
n
a
a
−
−
= ⋅ = ⋅ =
= ⋅ =
24. 5 1 4
5
1 1
1 1 11 1
3 3 81
1 11
3 3
n n
n
a
a
−
− −
= ⋅ − = ⋅ − =
= ⋅ − = −
25. ( ) ( )( ) ( )
5 1 4
5
1
2 2 2 2 2 4 4 2
2 2 2n n
n
a
a
−
−
= ⋅ = ⋅ = ⋅ =
= ⋅ =
26. 5 1 4
5
1
1 10 0 0
10 0
n
n
a
a
−
−
= ⋅ = ⋅ = π π
= ⋅ = π
27. 1
7 1 6
7
11, , 7
2
1 1 11
2 2 64
a r n
a−
= = =
= ⋅ = =
28. 1
8 1 78
1, 3, 8
1 3 3 2187
a r n
a −
= = =
= ⋅ = =
29.
( ) ( )1
9 1 89
1, 1, 9
1 1 1 1
a r n
a−
= = − =
= ⋅ − = − =
30.
( ) ( )1
10 1 910
1, 2, 10
1 2 1 2 1( 512) 512
a r n
a−
= − = − =
= − ⋅ − = − ⋅ − = − − =
31.
( ) ( )1
8 1 78
0.4, 0.1, 8
0.4 0.1 0.4 0.1 0.00000004
a r n
a−
= = =
= ⋅ = =
32.
( )1
67 17
0.1, 10, 7
0.1 10 0.1 10 100,000
a r n
a −
= = =
= ⋅ = =
33. 1 7a = , 14
27
r = = , 11
nna a r −= ⋅
17 2nna −= ⋅
34. 1 5a = , 10
25
r = = , 11
nna a r −= ⋅
15 2nna −= ⋅
35. 1 3a = − , 1 1
3 3r = = −
−, 1
1n
na a r −= ⋅
1 21 1
33 3
n n
na− −
= − − = −
36. 1 4a = , 1
4r = , 1
1n
na a r −= ⋅
1 21 1
44 4
n n
na− −
= =
Chapter 12: Sequences; Induction; the Binomial Theorem
1268
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
37. 11
nna a r −= ⋅
( )( )
6 1
1
5
1
1
1
243 3
243 3
243 243
1
a
a
a
a
−= ⋅ −
= −
= −− =
Therefore, 1( 3)nna −= − −
38. 11
nna a r −= ⋅
2 1
1
1
1
17
3
17
321
a
a
a
− =
=
=
Therefore, 1
121
3
n
na−
=
.
39. 4 1 3
24 12 1
2 1
a a r rr
a ra r
−
−⋅
= = =⋅
2 1575225
7
225 15
r
r
= =
= =
11
2 11
1
1
7 15
7 15
7
15
nna a r
a
a
a
−
−
= ⋅
= ⋅=
=
Therefore, 1 2715 7 15
15n n
na − −= ⋅ = ⋅ .
40. 6 1 5
36 13 1 2
3 1
a a r rr
a a r r
−
−⋅
= = =⋅
13 81
13
3
1 13
81 27
1 1
27 3
r
r
= = ⋅ =
= =
11
3 1
1
1
1
1 1
3 3
1 1
3 93
nna a r
a
a
a
−
−
= ⋅
= ⋅
=
=
Therefore, 1 2
1 13
3 3
n n
na− −
= =
.
41.
( )
( )
1
1
1, 2
4
1 1 1 2 11 2
1 4 1 2 4
12 1
4
n nn
n
n
a r
rS a
r
= =
− −= = = − − − −
= −
42.
( ) ( )
1
1
3 1, 3
9 3
1 1 1 3 1 1 3
1 3 1 3 3 2
1 11 3 3 1
6 6
n n n
n
n n
a r
rS a
r
= = =
− − −= = = − − −
= − − = −
43. 12 2
,3 3
a r= =
1
21
1 2 321 3 13
21
2 232 1
13 33
n
n
n
n
n
rS a
r
− − = = − −
− = = −
Section 12.3: Geometric Sequences; Geometric Series
1269
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
44.
( ) ( )
1
1
4, 3
1 1 3 1 34 4
1 1 3 2
2 1 3 2 3 1
n n n
n
n n
a r
rS a
r
= =
− − −= = = − − −
= − − = −
45. 1
1
1, 2
1 1 21 1 2
1 1 2
n nn
n
a r
rS a
r
= − =
− −= = − = − − −
46. 13
2,5
a r= =
1
3 31 1
1 5 52 2
3 21 15 5
35 1
5
n n
n
n
n
rS a
r
− − − = = = − −
= −
47. Using the sum of the sequence feature:
48. Using the sum of the sequence feature:
49. Using the sum of the sequence feature:
50. Using the sum of the sequence feature:
51. Using the sum of the sequence feature:
52. Using the sum of the sequence feature:
53. 1
1
11,
3Since 1, the series converges.
1 1 31 21 2
13 3
a r
r
aS
r∞
= =
<
= = = =− −
54. 1
1
22,
3Since 1, the series converges.
2 26
2 111
3 3
a r
r
aS
r∞
= =
<
= = = =− −
55. 1
1
18,
2Since 1, the series converges.
8 816
1 111
2 2
a r
r
aS
r∞
= =
<
= = = =− −
56. 1
1
16,
3Since 1, the series converges.
6 69
1 211
3 3
a r
r
aS
r∞
= =
<
= = = =− −
Chapter 12: Sequences; Induction; the Binomial Theorem
1270
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
57. 1
1
12,
4Since 1, the series converges.
2 2 851 51
144
a r
r
aS
r∞
= = −
<
= = = =− − −
58. 1
1
31,
4Since 1, the series converges.
1 1 471 73
144
a r
r
aS
r∞
= = −
<
= = = =− − −
59. 1 8a = , 3
2r =
Since 1r > , the series diverges.
60. 1 9a = , 4
3r =
Since 1r > , the series diverges.
61. 1
1
15,
4Since 1, the series converges.
5 5 201 31 3
14 4
a r
r
aS
r∞
= =
<
= = = =− −
62. 1
1
18,
3Since 1, the series converges.
8 812
1 211
3 3
a r
r
aS
r∞
= =
<
= = = =− −
63. 11
2a = , 3r =
Since 1r > , the series diverges.
64. 1 3a = , 3
2r =
Since 1r > , the series diverges.
65. 1
1
26,
3Since 1, the series converges.
6 6 1851 52
133
a r
r
aS
r∞
= = −
<
= = = =− − −
66. 1
1
14,
2Since 1, the series converges.
4 4 831 31
122
a r
r
aS
r∞
= = −
<
= = = =− − −
67. 1 1
1 1 1
2 2 2 23 3 2
3 3 3 3
k k k
k k k
− −∞ ∞ ∞
= = =
= ⋅ ⋅ =
1 2a = , 2
3r =
12 13 3
Since 1, the series converges.
2 26
1 1
r
aS
r∞
<
= = = =− −
68. 1 1
1 1 1
3 3 3 3 32 2
4 4 4 2 4
k k k
k k k
− −∞ ∞ ∞
= = =
= ⋅ ⋅ =
13
2a = ,
3
4r =
13214
Since 1, the series converges.
332 4 6
31 21 4
r
aS
r∞
<
= = = = ⋅ =− −
69. { }2n +
( 1 2) ( 2) 3 2 1d n n n n= + + − + = + − − =
The difference between consecutive terms is constant. Therefore the sequence is arithmetic.
50 50 50
501 1 1
( 2) 2
50(50 1)2(50) 1275 100 1375
2
k k k
S k k= = =
= + = +
+= + = + =
70. { }2 5n −
2( 1) 5 (2 5)
2 2 5 2 5 2
d n n
n n
= + − − −= + − − + =
The difference between consecutive terms is constant. Therefore the sequence is arithmetic.
Section 12.3: Geometric Sequences; Geometric Series
1271
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
50 50 50
501 1 1
(2 5) 2 5
50(50 1)2 5(50) 2550 250 2300
2
k k k
S k k= = =
= − = −
+ = − = − =
71. { }24n Examine the terms of the sequence: 4,
16, 36, 64, 100, ... There is no common difference and there is no common ratio. Therefore the sequence is neither arithmetic nor geometric.
72. { }25 1n + Examine the terms of the sequence:
6, 21, 46, 81, 126, ... There is no common difference and there is no common ratio. Therefore the sequence is neither arithmetic nor geometric.
73. 2
33
n −
2 23 ( 1) 3
3 3
2 2 2 23 3
3 3 3 3
d n n
n n
= − + − −
= − − − + = −
The difference between consecutive terms is constant. Therefore the sequence is arithmetic.
50 50 50
501 1 1
2 23 3
3 3
2 50(50 1)3(50) 150 850 700
3 2
k k k
S k k= = =
= − = −
+ = − = − = −
74. 3
84
n −
3 38 ( 1) 8
4 4
3 3 3 38 8
4 4 4 4
d n n
n n
= − + − −
= − − − + = −
The difference between consecutive terms is constant. Therefore the sequence is arithmetic.
50 50 50
501 1 1
3 38 8
4 4
3 50(50 1)8(50)
4 2
400 956.25 556.25
k k k
S k k= = =
= − = −
+ = −
= − = −
75. 1, 3, 6, 10, ... There is no common difference and there is no common ratio. Therefore the sequence is neither arithmetic nor geometric.
76. 2, 4, 6, 8, ... The common difference is 2. The difference between consecutive terms is constant. Therefore the sequence is arithmetic.
( )50 50
501 1
50(50 1)2 2 2 2550
2k k
S k k= =
+ = = = =
77. 2
3
n
1
12
2 233 32
3
n
n n
nr
+
+ − = = =
The ratio of consecutive terms is constant. Therefore the sequence is geometric.
50
50
501
21
2 2 31.999999997
23 3 13
k
k
S=
− = = ⋅ = −
78. 5
4
n
1
15
5 544 45
4
n
n n
nr
+
+ − = = =
The ratio of consecutive terms is constant. Therefore the sequence is geometric.
50
50
501
51
5 5 4350,319.6161
54 4 14
k
k
S=
− = = ⋅ ≈ −
79. –1, 2, –4, 8, ... 2 4 8
21 2 4
r−= = = = −
− −
The ratio of consecutive terms is constant. Therefore the sequence is geometric.
50501
501
14
1 ( 2)1 ( 2) 1
1 ( 2)
3.752999689 10
k
k
S −
=
− −= − ⋅ − = − ⋅− −
≈ ×
Chapter 12: Sequences; Induction; the Binomial Theorem
1272
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
80. 1, 1, 2, 3, 5, 8, ... There is no common difference and there is no common ratio. Therefore the sequence is neither arithmetic nor geometric.
81. { }/ 23n
112
1/ 22 2
2
33 3
3
nn n
nr
+ + −
= = =
The ratio of consecutive terms is constant. Therefore the sequence is geometric.
( )501/ 250/ 2 1/ 2
50 1/ 21
12
1 33 3
1 3
2.004706374 10
k
k
S=
−= = ⋅
−≈ ×
82. { }( 1)n−
11( 1)
( 1) 1( 1)
nn n
nr
++ −−= = − = −
−
The ratio of consecutive terms is constant. Therefore the sequence is geometric.
5050
501
1 ( 1)( 1) ( 1) 0
1 ( 1)k
k
S=
− −= − = − ⋅ =− −
83. Find the common ratio of the terms and solve the system of equations:
2 2
2 3;
22 3
4 4 3 42
x xr r
x xx x
x x x x xx x
+ += =+
+ += → + + = + → = −+
84. Find the common ratio of the terms and solve the system of equations:
2 2
2;
12
2 21
x xr r
x xx x
x x x xx x
+= =−+ = → + − = → =
−
85. This is a geometric series with
1 $18,000, 1.05, 5a r n= = = . Find the 5th
term:
( ) ( )5 1 45 18000 1.05 18000 1.05 $21,879.11a
−= = =
86. This is a geometric series with
1 $15,000, 0.85, 6a r n= = = . Find the 6th
term:
( ) ( )6 1 56 15000 0.85 15000 0.85 $6655.58a
−= = =
87. a. Find the 10th term of the geometric sequence:
1
10 1 910
2, 0.9, 10
2(0.9) 2(0.9) 0.775 feet
a r n
a −
= = =
= = =
b. Find when 1nn a < :
( )( ) ( )
( )( )( )( )
1
1
2(0.9) 1
0.9 0.5
( 1) log 0.9 log 0.5
log 0.51
log 0.9
log 0.51 7.58
log 0.9
n
n
n
n
n
−
−
<
<
− <
− >
> + ≈
On the 8th swing the arc is less than 1 foot.
c. Find the sum of the first 15 swings:
( )
( )( )
1515
15
15
1 0.91 (0.9)2 2
1 0.9 0.1
20 1 0.9 15.88 feet
S −− = = −
= − =
d. Find the infinite sum of the geometric series: 2 2
20 feet1 0.9 0.1
S∞ = = =−
88. a. Find the 3rd term of the geometric sequence:
1
3 1 23
24, 0.8, 3
24(0.8) 24(0.8) 15.36 feet
a r n
a −
= = =
= = =
b. The height after the n th bounce is:
( ) ( )( )
1124(0.8) 24 0.8 0.8
30 0.8 ft
nnn
n
a−−= =
=
c. Find when 0.5nn a < :
( )( ) ( )
( )( )
( )( )
1
1
24(0.8) 0.5
0.8 0.020833
( 1) log 0.8 log 0.020833
log 0.0208331
log 0.8
log 0.0208331 18.35
log 0.8
n
n
n
n
n
−
−
<
<
− <
− >
> + ≈
On the 19th bounce the height is less than 0.5 feet.
Section 12.3: Geometric Sequences; Geometric Series
1273
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
d. Find the infinite sum of the geometric series:
24 24120 feet
1 0.8 0.2S∞ = = =
− on the upward
bounce.
For the downward motion of the ball: 30 30
150 feet1 0.8 0.2
S∞ = = =−
The total distance the ball travels is 120 + 150 = 270 feet.
89. This is an ordinary annuity with $100P = and
( ) ( )12 30 360n = = payment periods. The
interest rate per period is 0.12
0.0112
= . Thus,
[ ]3601 0.01 1
100 $349,496.410.01
A + − = ≈
90. This is an ordinary annuity with $400P = and
( )( )12 3 36n = = payment periods. The interest
rate per period is 0.10
12. Thus,
360.10
1 112
400 $16,712.730.1012
A
+ − = ≈
91. This is an ordinary annuity with $500P = and
( )( )4 20 80n = = payment periods. The interest
rate per period is 0.08
0.024
= . Thus,
[ ]801 0.02 1
500 $96,885.980.02
A + − = ≈
92. This is an ordinary annuity with $1000P = and
( )( )2 15 30n = = payment periods. The interest
rate per period is 0.10
0.052
= . Thus,
[ ]301 0.05 1
1000 $66,438.850.05
A + − = ≈
93. This is an ordinary annuity with $50,000A =
and ( )( )12 10 120n = = payment periods. The
interest rate per period is 0.06
0.00512
= . Thus,
[ ]
[ ]
120
120
1 0.005 150,000
0.005
0.00550,000 $305.10
1 0.005 1
P
P
+ − = = ≈ + −
94. This is an ordinary annuity with $150,000A =
and ( )( )12 18 216n = = payment periods. The
interest rate per period is 0.08
12. Thus,
216
216
0.081 1
12150,000
0.0812
0.0812150,000 $312.44
0.081 1
12
P
P
+ − = = ≈ + −
95. This is a geometric sequence with
1 1, 2, 64a r n= = = .
Find the sum of the geometric series: 64 64
6464
19
1 2 1 21 2 1
1 2 1
1.845 10 grains
S − −= = = − − −
= ×
96. This is an infinite geometric series with
11 1,4 4a r= = .
Find the sum of the infinite geometric series:
( )( )
( )( )
1 114 4
1 3 31 4 4
S∞ = = =−
1 3
∴ of the square is eventually shaded.
97. The common ratio, 0.90 1r = < . The sum is: 1 1
101 0.9 0.10
S = = =−
.
The multiplier is 10.
Chapter 12: Sequences; Induction; the Binomial Theorem
1274
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
98. The common ratio, 0.95 1r = < . The sum is: 1 1
201 0.95 0.05
S = = =−
.
The multiplier is 20.
99. This is an infinite geometric series with 1.03
4, and 1.09
a r= = .
Find the sum: 4
Price $72.671.03
11.09
= ≈ −
.
100. This is an infinite geometric series with
11.04
2.5, and 1.11
a r= = .
Find the sum: 2.5
Price $39.641.04
11.11
= ≈ −
.
101. Given: 1 1000, 0.9a r= =
Find when 0.01nn a < :
( )( ) ( )
( )( )
( )( )
1
1
1000(0.9) 0.01
0.9 0.00001
( 1) log 0.9 log 0.00001
log 0.000011
log 0.9
log 0.000011 110.27
log 0.9
n
n
n
n
n
−
−
<
<
− <
− >
> + ≈
On the 111th day or December 20, 2007, the amount will be less than $0.01.
Find the sum of the geometric series:
( )
( )
111
111 1
111
1 0.911000
1 1 0.9
1 0.91000 $9999.92
0.1
nrS a
r
−− = = − − − = =
102. Both options are geometric sequences: Option A: 1 $20,000; 1.06; 5a r n= = =
( )
5 1 45
5
5
20,000(1.06) 20,000(1.06) $25, 250
1 1.0620000 $112,742
1 1.06
a
S
−= = =
− = = −
Option B: 1 $22,000; 1.03; 5a r n= = =
( )
5 1 45
5
5
22,000(1.03) 22,000(1.03) $24,761
1 1.0322000 $116,801
1 1.03
a
S
−= = =
− = = −
Option A provides more money in the 5th year, while Option B provides the greatest total amount of money over the 5 year period.
103. Find the sum of each sequence: A: Arithmetic series with:
1 $1000, 1, 1000a d n= = − =
Find the sum of the arithmetic series:
( )10001000
1000 1 500(1001) $500,5002
S = + = =
B: This is a geometric sequence with 1 1, 2, 19a r n= = = .
Find the sum of the geometric series: 19 19
1919
1 2 1 21 2 1 $524, 287
1 2 1S
− −= = = − = − −
B results in more money.
104. Option 1: Total Salary $2,000,000(7) $100,000(7)
$14,700,000
= +=
Option 2: Geometric series with:
1 $2,000,000, 1.045, 7a r n= = =
Find the sum of the geometric series:
( )71 1.045
2,000,000 $16,038,3041 1.045
S − = ≈ −
Option 3: Arithmetic series with:
1 $2,000,000, $95,000, 7a d n= = =
Find the sum of the arithmetic series:
( )77
2(2,000,000) (7 1)(95,000)2$15,995,000
S = + −
=
Option 2 provides the most money; Option 1 provides the least money.
105. The amount paid each day forms a geometric sequence with 1 0.01a = and 2r = .
22 22
22 11 1 2
0.01 41,943.031 1 2
rS a
r
− −= ⋅ = ⋅ =− −
Section 12.4: Mathematical Induction
1275
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
The total payment would be $41,943.03 if you worked all 22 days.
( )2122 122 1 0.01 2 20,971.52a a r −= ⋅ = =
The payment on the 22nd day is $20,971.52.
Answers will vary. With this payment plan, the bulk of the payment is at the end so missing even one day can dramatically reduce the overall payment. Notice that with one sick day you would lose the amount paid on the 22nd day which is about half the total payment for the 22 days.
106. Yes, a sequence can be both arithmetic and geometric. For example, the constant sequence 3,3,3,3,..... can be viewed as an arithmetic
sequence with 1 3a = and 0.d = Alternatively,
the same sequence can be viewed as a geometric sequence with 1 3a = and 1.r =
107. Answers will vary.
108. Answers will vary.
109. Answers will vary. Both increase (or decrease) exponentially, but the domain of a geometric sequence is the set of natural numbers while the domain of an exponential function is the set of all real numbers.
Section 12.4
1. I: 1: 2 1 2 and 1(1 1) 2n = ⋅ = + =
II: If 2 4 6 2 ( 1)k k k+ + + + = + , then
[ ]
( ) ( )( )
2 4 6 2 2( 1)
2 4 6 2 2( 1)
( 1) 2( 1)
( 1)( 2)
1 1 1
k k
k k
k k k
k k
k k
+ + + + + += + + + + + += + + += + +
= + + +
Conditions I and II are satisfied; the statement is true.
2. I: 1: 4 1 3 1 and 1(2 1 1) 1n = ⋅ − = ⋅ − =
II: If 1 5 9 (4 3) (2 1)k k k+ + + + − = − , then
[ ]
( ) ( )( )
2
2
1 5 9 (4 3) (4( 1) 3)
1 5 9 (4 3) 4 4 3
(2 1) 4 1
2 4 1
2 3 1
( 1)(2 1)
1 2 1 1
k k
k k
k k k
k k k
k k
k k
k k
+ + + + − + + −= + + + + − + + −= − + +
= − + +
= + += + +
= + + −
Conditions I and II are satisfied; the statement is true.
3. I: 1
1: 1 2 3 and 1(1 5) 32
n = + = ⋅ + =
II: If 1
3 4 5 ( 2) ( 5)2
k k k+ + + + + = ⋅ + , then
[ ]
( )
( )( )
2
2
2
3 4 5 ( 2) [( 1) 2]
3 4 5 ( 2) ( 3)
1( 5) ( 3)
21 5
32 21 7
32 21
7 621
( 1)( 6)21
( 1) 1 52
k k
k k
k k k
k k k
k k
k k
k k
k k
+ + + + + + + += + + + + + + +
= ⋅ + + +
= + + +
= + +
= ⋅ + +
= ⋅ + +
= ⋅ + + +
Conditions I and II are satisfied; the statement is true.
4. I: 1: 2 1 1 3 and 1(1 2) 3n = ⋅ + = + =
II: If 3 5 7 (2 1) ( 2)k k k+ + + + + = + , then
[ ]
( )( )
2
2
3 5 7 (2 1) [2( 1) 1]
3 5 7 (2 1) (2 3)
( 2) (2 3)
2 2 3
4 3
( 1)( 3)
( 1) 1 2
k k
k k
k k k
k k k
k k
k k
k k
+ + + + + + + += + + + + + + += + + +
= + + +
= + += + +
= + + +
Conditions I and II are satisfied; the statement is true.
Chapter 12: Sequences; Induction; the Binomial Theorem
1276
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
5. I: 1
1: 3 1 1 2 and 1(3 1 1) 22
n = ⋅ − = ⋅ ⋅ + =
II: If 1
2 5 8 (3 1) (3 1)2
k k k+ + + + − = ⋅ + ,
then
[ ]
( )
( )( )
2
2 2
2 5 8 (3 1) [3( 1) 1]
2 5 8 (3 1) (3 2)
1 3 1(3 1) (3 2) 3 2
2 2 23 7 1
2 3 7 42 2 21
( 1)(3 4)21
( 1) 3 1 12
k k
k k
k k k k k k
k k k k
k k
k k
+ + + + − + + −= + + + + − + +
= ⋅ + + + = + + +
= + + = ⋅ + +
= ⋅ + +
= ⋅ + + +
Conditions I and II are satisfied; the statement is
true.
6. I: 1
1: 3 1 2 1 and 1(3 1 1) 12
n = ⋅ − = ⋅ ⋅ − =
II: If 1
1 4 7 (3 2) (3 1)2
k k k+ + + + − = ⋅ − ,
then
[ ]
( )
( )( )
2
2 2
1 4 7 (3 2) [3( 1) 2]
1 4 7 (3 2) (3 1)
1 3 1(3 1) (3 1) 3 1
2 2 23 5 1
1 3 5 22 2 21
( 1)(3 2)21
( 1) 3 1 12
k k
k k
k k k k k k
k k k k
k k
k k
+ + + + − + + −= + + + + − + +
= ⋅ − + + = − + +
= + + = ⋅ + +
= ⋅ + +
= ⋅ + + −
Conditions I and II are satisfied; the statement is
true.
7. I: 1 1 11: 2 1 and 2 1 1n −= = − =
II: If 2 11 2 2 2 2 1k k−+ + + + = − , then 2 1 1 1
2 1
1
1 2 2 2 2
1 2 2 2 2
2 1 2 2 2 1
2 1
k k
k k
k k k
k
− + −
−
+
+ + + + +
= + + + + +
= − + = ⋅ −
= −
Conditions I and II are satisfied; the statement is true.
8. I: 1 1 111: 3 1 and (3 1) 1
2n −= = − =
II: If 2 1 11 3 3 3 (3 1)
2k k−+ + + + = ⋅ − , then
( )( )
2 1 1 1
2 1
1
1 3 3 3 3
1 3 3 3 3
1 1 1(3 1) 3 3 3
2 2 23 1 1
3 3 3 12 2 21
3 12
k k
k k
k k k k
k k
k
− + −
−
+
+ + + + +
= + + + + +
= ⋅ − + = ⋅ − +
= ⋅ − = ⋅ ⋅ −
= −
Conditions I and II are satisfied; the statement is true.
9. I: ( )1 1 111: 4 1 and 4 1 1
3n −= = ⋅ − =
II: If ( )2 1 11 4 4 4 4 1
3k k−+ + + + = ⋅ − , then
( )( )
( )
2 1 1 1
2 1
1
1 4 4 4 4
1 4 4 4 4
1 1 14 1 4 4 4
3 3 34 1 1
4 4 4 13 3 31
4 13
k k
k k
k k k k
k k
k
− + −
−
+
+ + + + +
= + + + + +
= ⋅ − + = ⋅ − +
= ⋅ − = ⋅ −
= ⋅ −
Conditions I and II are satisfied; the statement is true.
10. I: ( )1 1 111: 5 1 and 5 1 1
4n −= = ⋅ − =
II: If ( )2 1 11 5 5 5 5 1
4k k−+ + + + = ⋅ − , then
( )( )
( )
2 1 1 1
2 1
1
1 5 5 5 5
1 5 5 5 5
1 1 15 1 5 5 5
4 4 45 1 1
5 5 5 14 4 41
5 14
k k
k k
k k k k
k k
k
− + −
−
+
+ + + + +
= + + + + +
= ⋅ − + = ⋅ − +
= ⋅ − = ⋅ −
= ⋅ −
Conditions I and II are satisfied; the statement is true.
Section 12.4: Mathematical Induction
1277
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
11. I: 1 1 1 1
1: and1(1 1) 2 1 1 2
n = = =+ +
II: If 1 1 1 1
1 2 2 3 3 4 ( 1) 1
k
k k k+ + + + =
⋅ ⋅ ⋅ + + , then
( )2
1 1 1 1 1 1 1 1 1 1
1 2 2 3 3 4 ( 1) ( 1)( 1 1) 1 2 2 3 3 4 ( 1) ( 1)( 2)
1 2 1
1 ( 1)( 2) 1 2 ( 1)( 2)
2 1 ( 1)( 1) 1 1
( 1)( 2) ( 1)( 2) 2 1 1
k k k k k k k k
k k k
k k k k k k k
k k k k k k
k k k k k k
+ + + + + = + + + + + ⋅ ⋅ ⋅ + + + + ⋅ ⋅ ⋅ + + +
+= + = ⋅ ++ + + + + + +
+ + + + + += = = =+ + + + + + +
Conditions I and II are satisfied; the statement is true.
12. I: 1 1 1 1
1: and(2 1 1)(2 1 1) 3 2 1 1 3
n = = =⋅ − ⋅ + ⋅ +
II: If 1 1 1 1
1 3 3 5 5 7 (2 1)(2 1) 2 1
k
k k k+ + + + =
⋅ ⋅ ⋅ − + + , then
2
1 1 1 1 1
1 3 3 5 5 7 (2 1)(2 1) (2( 1) 1)(2( 1) 1)
1 1 1 1 1
1 3 3 5 5 7 (2 1)(2 1) (2 1)(2 3)
1 2 3 1
2 1 (2 1)(2 3) 2 1 2 3 (2 1)(2 3)
2 3 1 ( 1)(2 1)
(2 1)(2 3) (2 1)(2 3)
k k k k
k k k k
k k k
k k k k k k k
k k k k k
k k k k
+ + + + +⋅ ⋅ ⋅ − + + − + +
= + + + + + ⋅ ⋅ ⋅ − + + +
+= + = ⋅ ++ + + + + + +
+ + + + += = =+ + + +
( )1 1
2 3 2 1 1
k
k k
+=+ + +
Conditions I and II are satisfied; the statement is true.
13. I: 2 11: 1 1 and 1(1 1)(2 1 1) 1
6n = = ⋅ + ⋅ + =
II: If 2 2 2 2 11 2 3 ( 1)(2 1)
6k k k k+ + + + = ⋅ + + , then
( )( ) ( )( )
2 2 2 2 2 2 2 2 2 2 2
2 2 2
11 2 3 ( 1) 1 2 3 ( 1) ( 1)(2 1) ( 1)
61 1 1 1 7 1
( 1) (2 1) 1 ( 1) 1 ( 1) 1 ( 1) 2 7 66 3 6 3 6 6
1( 1)( 2)(2 3)
61
( 1) 1 1 2 1 16
k k k k k k k k
k k k k k k k k k k k k k k
k k k
k k k
+ + + + + + = + + + + + + = + + + +
= + + + + = + + + + = + + + = + + +
= ⋅ + + +
= ⋅ + + + + +
Conditions I and II are satisfied; the statement is true.
Chapter 12: Sequences; Induction; the Binomial Theorem
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14. I: 3 2 211: 1 1 and 1 (1 1) 1
4n = = ⋅ + =
II: If 3 3 3 3 2 211 2 3 ( 1)
4k k k+ + + + = + , then
3 3 3 3 3 3 3 3 3 3 2 2 3
2 2 2 2
2 2
2 2
11 2 3 ( 1) 1 2 3 ( 1) ( 1) ( 1)
41 1
( 1) 1 ( 1) 4 44 4
1( 1) ( 2)
41
( 1) (( 1) 1)4
k k k k k k k
k k k k k k
k k
k k
+ + + + + + = + + + + + + = + + +
= + + + = + + +
= ⋅ + +
= ⋅ + + +
Conditions I and II are satisfied; the statement is true.
15. I: 1
1: 5 1 4 and 1(9 1) 42
n = − = ⋅ − =
II: If 1
4 3 2 (5 ) (9 )2
k k k+ + + + − = ⋅ − , then
( ) [ ]
[ ]
2 2 2
14 3 2 (5 ) 5 ( 1) 4 3 2 (5 ) (4 ) (9 ) (4 )
29 1 1 7 1
4 4 7 82 2 2 2 2
1 1 1( 1)( 8) ( 1)(8 ) ( 1) 9 ( 1)
2 2 2
k k k k k k k
k k k k k k k
k k k k k k
+ + + + − + − + = + + + + − + − = − + −
= − + − = − + + = − ⋅ − −
= − ⋅ + − = ⋅ + − = ⋅ + − +
Conditions I and II are satisfied; the statement is true.
16. I: 1
1: (1 1) 2 and 1(1 3) 22
n = − + = − − ⋅ + = −
II: If 1
2 3 4 ( 1) ( 3)2
k k k− − − − − + = − ⋅ + , then
( ) [ ]2 2
2
2 3 4 ( 1) ( 1) 1 2 3 4 ( 1) ( 2)
1 1 3 1 5( 3) ( 2) 2 2
2 2 2 2 21 1
5 4 ( 1)( 4)2 21
( 1)(( 1) 3)2
k k k k
k k k k k k k k
k k k k
k k
− − − − − + − + + = − − − − − + − +
= − ⋅ + − + = − − − − = − − −
= − ⋅ + + = − ⋅ + +
= − ⋅ + + +
Conditions I and II are satisfied; the statement is true.
Section 12.4: Mathematical Induction
1279
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
17. I: 1
1: 1(1 1) 2 and 1(1 1)(1 2) 23
n = + = ⋅ + + =
II: If 1
1 2 2 3 3 4 ( 1) ( 1)( 2)3
k k k k k⋅ + ⋅ + ⋅ + + + = ⋅ + + , then
[ ]1 2 2 3 3 4 ( 1) ( 1)( 1 1) 1 2 2 3 3 4 ( 1) ( 1)( 2)
1 1( 1)( 2) ( 1)( 2) ( 1)( 2) 1
3 3
1( 1)( 2)( 3)
31
( 1)(( 1) 1)(( 1) 2)3
k k k k k k k k
k k k k k k k k
k k k
k k k
⋅ + ⋅ + ⋅ + + + + + + + = ⋅ + ⋅ + ⋅ + + + + + +
= ⋅ + + + + + = + + +
= ⋅ + + +
= ⋅ + + + + +
Conditions I and II are satisfied; the statement is true.
18. I: 1
1: (2 1 1)(2 1) 2 and 1(1 1)(4 1 1) 23
n = ⋅ − ⋅ = ⋅ + ⋅ − =
II: If 1
1 2 3 4 5 6 (2 1)(2 ) ( 1)(4 1)3
k k k k k⋅ + ⋅ + ⋅ + + − = ⋅ + − , then
[ ]
( )2 2
2
1 2 3 4 5 6 (2 1)(2 ) (2( 1) 1)(2( 1))
1 2 3 4 5 6 (2 1)(2 ) (2 1)( 1) 2
1 1( 1)(4 1) 2( 1)(2 1) ( 1) (4 1) 2(2 1)
3 3
4 1 1( 1) 4 2 ( 1) 4 12 6
3 3 3
1( 1) 4 11
3
k k k k
k k k k
k k k k k k k k k
k k k k k k k k
k k k
⋅ + ⋅ + ⋅ + + − + + − += ⋅ + ⋅ + ⋅ + + − + + + ⋅
= + − + + + = + ⋅ − + + = + − + + = ⋅ + − + +
= + +
( ) 16 ( 1)( 2)(4 3)
31
( 1)(( 1) 1)(4( 1) 1)3
k k k
k k k
+ = ⋅ + + +
= ⋅ + + + + −
Conditions I and II are satisfied; the statement is true.
19. I: 21: 1 1 2 is divisible by 2n = + =
II: If 2 is divisible by 2k k+ , then 2 2
2
( 1) ( 1) 2 1 1
( ) (2 2)
k k k k k
k k k
+ + + = + + + +
= + + +
Since 2k k+ is divisible by 2 and 2 2k + is
divisible by 2, then 2( 1) ( 1)k k+ + + is
divisible by 2.
Conditions I and II are satisfied; the statement is true.
20. I: 31: 1 2 1 3 is divisible by 3n = + ⋅ =
II: If 3 2 is divisible by 3k k+ , then 3
3 2
3 2
( 1) 2( 1)
3 3 1 2 2
( 2 ) (3 3 3)
k k
k k k k
k k k k
+ + +
= + + + + +
= + + + +
Since 3 2k k+ is divisible by 3 and 23 3 3k k+ + is divisible by 3, then
3( 1) 2( 1)k k+ + + is divisible by 3.
Conditions I and II are satisfied; the statement is true.
Chapter 12: Sequences; Induction; the Binomial Theorem
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
21. I: 21: 1 1 2 2 is divisible by 2n = − + =
II: If 2 2 is divisible by 2k k− + , then 2 2
2
( 1) ( 1) 2 2 1 1 2
( 2) (2 )
k k k k k
k k k
+ − + + = + + − − +
= − + +
Since 2 2k k− + is divisible by 2 and 2k is
divisible by 2, then 2( 1) ( 1) 2k k+ − + + is
divisible by 2.
Conditions I and II are satisfied; the statement is true.
22. I: 1: 1(1 1)(1 2) 6 is divisible by 6n = + + =
II: If ( 1)( 2) is divisible by 6k k k+ + , then
( 1)( 1 1)( 1 2)
( 1)( 2)( 3)
( 1)( 2) 3( 1)( 2).
k k k
k k k
k k k k k
+ + + + += + + += + + + + +
Now, ( 1)( 2) is divisible by 6;
and since either 1 or 2 is even,
k k k
k k
+ ++ +
( )( )3 1 2k k+ + is divisible by 6.
Thus, ( )( )( )( )( ) ( ) ( )1 2 3
1 2 3 1 2
k k k
k k k k k
+ + +
= + + + + +
is divisible by 6.
Conditions I and II are satisfied; the statement is true.
23. I: 11: If 1 then 1.n x x x= > = >
II: Assume, for some natural number k, that if
1x > , then 1kx > . 1
1
Then 1, for 1,
1 1
( 1)
k
k k
k
x x
x x x x x
x
+
+
> >
= ⋅ > ⋅ = >↑
>
Conditions I and II are satisfied; the statement is true.
24. I: 11: If 0 1 then 0 1.n x x= < < < <
II: Assume, for some natural number k, that if
0 1x< < , then 0 1kx< < .
1
1
Then, for 0 1,
0 1 1
Thus, 0 1.
k k
k
x
x x x x x
x
+
+
< <
< = ⋅ < ⋅ = <
< <
Conditions I and II are satisfied; the statement is true.
25. I: 1 11: is a factor of .n a b a b a b= − − = −
II: If is a factor of k ka b a b− − , show that
a b− is a factor of 1 1k ka b+ +− .
( )
1 1
( )
k k k k
k k k k
k k k
a b a a b b
a a a b a b b b
a a b b a b
+ +− = ⋅ − ⋅
= ⋅ − ⋅ + ⋅ − ⋅
= − + −
Since a b− is a factor of k ka b− and a b− is a factor of a b− , then a b− is a factor of
1 1k ka b+ +− .
Conditions I and II are satisfied; the statement is true.
26. I:
( ) ( )2 1 1 2 1 1 3 3
3 3 2 2
1:
is a factor of .
n
a b a b a b
a b a b a ab b
⋅ + ⋅ +
=
+ + = +
+ = + − +
II: 2 1 2 1If is a factor of k ka b a b+ ++ + , show that a b+ is a factor of
( ) ( )2 1 1 2 1 1k ka b+ + + ++ .
( )
2( 1) 1 2( 1) 1 2 3 2 3
2 2 1 2 2 1 2 2 1 2 2 1
2 2 1 2 1 2 1 2 2( )
k k k k
k k k k
k k k
a b a b
a a a b a b b b
a a b b a b
+ + + + + +
+ + + +
+ + +
+ = +
= ⋅ + ⋅ − ⋅ + ⋅
= + − −
Since a b+ is a factor of 2 1 2 1k ka b+ ++ and
a b+ is a factor of 2 2a b− ( )( )[ ]a b a b− + ,
then a b+ is a factor of 2 3 2 3k ka b+ ++ .
Conditions I and II are satisfied; the statement is true.
27. I: 1n = : ( )11 1 1 1a a a+ = + ≥ + ⋅
II: Assume that there is an integer k for which the inequality holds. We need to show that if
( )1 1k
a ka+ ≥ + then
( ) ( )11 1 1
ka k a
++ ≥ + + .
( ) ( ) ( )( )( )
( )( )
1
2
2
1 1 1
1 1
1
1 1
1 1
k ka a a
ka a
ka a ka
k a ka
k a
++ = + +
≥ + +
= + + +
= + + +
≥ + +
Conditions I and II are satisfied, the statement is true.
Section 12.4: Mathematical Induction
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28. 2
1:
1 1 41 41 is a prime number.
n =
− + =
2 2
41:
41 41 41 41 is not a prime number.
n =
− + =
29. II: If 22 4 6 2 2k k k+ + + + = + + , then
[ ]2
2
2
2 4 6 2 2( 1)
2 4 6 2 2 2
2 2 2
( 2 1) ( 1) 2
( 1) ( 1) 2
k k
k k
k k k
k k k
k k
+ + + + + += + + + + + +
= + + + +
= + + + + +
= + + + +
I: 21: 2 1 2 and 1 1 2 4 2n = ⋅ = + + = ≠
30. I: 1
1 1 11: and
1
rn a r a a a
r− −= = = −
II: If 2 1 1
1
kk r
a a r a r a r ar
− −+ + + + = − ,
then 2 1 1 1
2 1
1
1
1
1
(1 ) (1 )
1
1
1
1
k k
k k
kk
k k
k k k
k
a a r a r a r a r
a a r a r a r a r
ra a r
r
a r a r r
r
a a r a r a r
r
ra
r
− + −
−
+
+
+ + + + +
= + + + + + −= + −
− + −=−
− + −=−
−= −
Conditions I and II are satisfied; the statement is true.
31. I: 1:
1(1 1)(1 1) and 1
2
n
a d a a d a
=−+ − = ⋅ + =
II: If [ ]( ) ( 2 ) ( 1)
( 1)
2
a a d a d a k d
k kka d
+ + + + + + + −−= +
then
[ ] ( )[ ][ ]
( ) ( ) ( )( )
2
2
( ) ( 2 ) ( 1)
( ) ( 2 ) ( 1) ( )
( 1)( )
2( 1)
( 1)2
2( 1)
2
( 1)2
( 1)( 1)
2
1 1 11
2
a a d a d a k d a kd
a a d a d a k d a kd
k kka d a kd
k kk a d k
k k kk a d
k kk a d
k kk a d
k kk a d
+ + + + + + + − + +
= + + + + + + + − + +
−= + + +
− = + + + − += + + += + +
+ = + + + + −
= + +
Conditions I and II are satisfied; the statement is true.
32. I: 4 :n = The number of diagonals of a
quadrilateral is 1
4(4 3) 22
⋅ − = .
II: Assume that for any integer k, the number of diagonals of a convex polygon with k sides
(k vertices) is 1
( 3)2
k k⋅ − . A convex
polygon with 1k + sides ( 1k + vertices) consists of a convex polygon with k sides (k vertices) plus a triangle, for a total of ( 1k + ) vertices. The diagonals of this
1k + -sided convex polygon consist of the diagonals of the k-sided polygon plus 1k − additional diagonals. For example, consider the following diagrams.
k = 5 sides
Chapter 12: Sequences; Induction; the Binomial Theorem
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k − 1 = 4 new diagonalsk + 1 = 6 sides
Thus, we have the equation:
( )
2
2
2
1 1 3( 3) ( 1) 1
2 2 21 1
12 21
221
( 1)( 2)21
( 1)(( 1) 3)2
k k k k k k
k k
k k
k k
k k
⋅ − + − = − + −
= − −
= ⋅ − −
= ⋅ + −
= ⋅ + + −
Conditions I and II are satisfied; the statement is true.
33. I: 3 :n = (3 2) 180 180 − ⋅ ° = ° which is the
sum of the angles of a triangle.
II: Assume that for any integer k, the sum of the angles of a convex polygon with k sides is ( 2) 180k − ⋅ ° . A convex polygon with
1k + sides consists of a convex polygon with k sides plus a triangle. Thus, the sum of the angles is ( 2) 180 180 (( 1) 2) 180 .k k− ⋅ ° + ° = + − ⋅ °
Conditions I and II are satisfied; the statement is true.
34. Answers will vary.
Section 12.5
1. Pascal Triangle
2. ! !
10 0! ( 0)! 1 !
n n n
n n
= = = − ⋅
! ( 1)!
1 1! ( 1)! 1 ( 1)!
n n n nn
n n
−= = = − ⋅ −
3. False; ( )
!
! !
n n
j j n j
= −
4. Binomial Theorem
5. 5 5! 5 4 3 2 1 5 4
103 3! 2! 3 2 1 2 1 2 1
⋅ ⋅ ⋅ ⋅ ⋅= = = = ⋅ ⋅ ⋅ ⋅ ⋅
6. 7 7! 7 6 5 4 3 2 1 7 6 5
353 3! 4! 3 2 1 4 3 2 1 3 2 1
⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅= = = = ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅
7. 7 7! 7 6 5 4 3 2 1 7 6
215 5! 2! 5 4 3 2 1 2 1 2 1
⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅= = = = ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅
8. 9 9! 9 8 7 6 5 4 3 2 1 9 8
367 7! 2! 7 6 5 4 3 2 1 2 1 2 1
⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅= = = = ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅
9. 50 50! 50 49! 50
5049 49!1! 49! 1 1
⋅= = = = ⋅
10. 100 100! 100 99 98! 100 99
495098 98! 2! 98! 2 1 2 1
⋅ ⋅ ⋅= = = = ⋅ ⋅ ⋅
11. 1000 1000! 1
11000 1000!0! 1
= = =
12. 1000 1000! 1
10 0!1000! 1
= = =
13. 1555 55!1.8664 10
23 23!32!
= ≈ ×
14. 1560 60!4.1918 10
20 20! 40!
= ≈ ×
15. 1347 47!1.4834 10
25 25! 22!
= ≈ ×
16. 1037 37!1.7673 10
19 19!18!
= ≈ ×
Section 12.5: The Binomial Theorem
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17. 5 5 4 3 2 1 05 5 5 5 5 5( 1)
0 1 2 3 4 5x x x x x x x
+ = + + + + +
5 4 3 25 10 10 5 1x x x x x= + + + + +
18. 5 5 4 2 3 3 2 4 1 5 0
5 4 3 2
5 5 5 5 5 5( 1) ( 1) ( 1) ( 1) ( 1) ( 1)
0 1 2 3 4 5
5 10 10 5 1
x x x x x x x
x x x x x
− = + − + − + − + − + −
= − + − + −
19. 6 6 5 4 2 3 3 2 4 5 0 6
6 5 4 3 2
6 5 4 3 2
6 6 6 6 6 6 6( 2) ( 2) ( 2) ( 2) ( 2) ( 2) ( 2)
0 1 2 3 4 5 6
6 ( 2) 15 4 20 ( 8) 15 16 6 ( 32) 64
12 60 160 240 192 64
x x x x x x x x
x x x x x x
x x x x x x
− = + − + − + − + − + − + −
= + − + ⋅ + − + ⋅ + ⋅ − +
= − + − + − +
20. 5 5 4 3 2 2 3 1 4 0 5
5 4 3 2
5 4 3 2
5 5 5 5 5 5( 3) (3) (3) (3) (3) (3)
0 1 2 3 4 5
5 (3) 10 9 10 (27) 5 81 243
15 90 270 405 243
x x x x x x x
x x x x x
x x x x x
+ = + + + + +
= + + ⋅ + + ⋅ +
= + + + + +
21. 4 4 3 2
4 3 2 4 3 2
4 4 4 4 4(3 1) (3 ) (3 ) (3 ) (3 )
0 1 2 3 4
81 4 27 6 9 4 3 1 81 108 54 12 1
x x x x x
x x x x x x x x
+ = + + + +
= + ⋅ + ⋅ + ⋅ + = + + + +
22. 5 5 4 3 2 2 3 4 5
5 4 3 2
5 4 3 2
5 5 5 5 5 5(2 3) (2 ) (2 ) 3 (2 ) 3 (2 ) 3 2 3 3
0 1 2 3 4 5
32 5 16 3 10 8 9 10 4 27 5 2 81 243
32 240 720 1080 810 243
x x x x x x
x x x x x
x x x x x
+ = + ⋅ + ⋅ + ⋅ + ⋅ ⋅ + ⋅
= + ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ +
= + + + + +
23. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )5 5 4 3 2 2 3 4 52 2 2 2 2 2 2 2 2 2 2 2
10 8 2 6 4 4 6 2 8 10
5 5 5 5 5 5
0 1 2 3 4 5
5 10 10 5
x y x x y x y x y x y y
x x y x y x y x y y
+ = + + + + +
= + + + + +
24. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( )
6 6 5 4 2 3 3 2 42 2 2 2 2 2 2 2 2 2 2
5 62 2 2
12 10 2 8 4 6 6 4 8 2 10 12
6 6 6 6 6
0 1 2 3 4
6 6
5 6
6 15 20 15 6
x y x x y x y x y x y
x y y
x x y x y x y x y x y y
− = + − + − + − + −
+ − + −
= − + − + − +
25. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( )( ) ( )
6 6 5 1 4 2 3 3
2 4 5 6
3 5/ 2 2 3/ 2 1/ 2
3 5/ 2 2 3/ 2 1/ 2
6 6 6 62 2 2 2
0 1 2 3
6 6 6 2 2 2
4 5 6
6 2 15 2 20 2 2 15 4 6 4 2 8
6 2 30 40 2 60 24 2 8
x x x x x
x x
x x x x x x
x x x x x x
+ = + + +
+ + +
= + + ⋅ + ⋅ + ⋅ + ⋅ +
= + + + + + +
Chapter 12: Sequences; Induction; the Binomial Theorem
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26. ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( )4 4 3 1 2 2 3 4
2 3/ 2 1/ 2
2 3/ 2 1/ 2
4 4 4 4 43 3 3 3 3
0 1 2 3 4
4 3 6 3 4 3 3 9
4 3 18 12 3 9
x x x x x
x x x x
x x x x
− = + − + − + − + −
= − + ⋅ − ⋅ +
= − + − +
27. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )5 5 4 3 2 2 3 4 5
5 5 4 4 3 3 2 2 2 2 3 3 4 4 5 5
5 5 5 5 5 5
0 1 2 3 4 5
5 10 10 5
ax by ax ax by ax by ax by ax by by
a x a x by a x b y a x b y axb y b y
+ = + ⋅ + + + +
= + + + + +
28. ( ) ( ) ( ) ( ) ( ) ( )( ) ( )4 4 3 2 2 3 4
4 4 3 3 2 2 2 2 3 3 4 4
4 4 4 4 4( )
0 1 2 3 4
4 6 4
ax by ax ax by ax by ax by by
a x a x by a x b y axb y b y
− = + − + − + − + −
= − + − +
29. 10, 4, , 3n j x x a= = = =
6 4 6 6
6
10 10! 10 9 8 73 81 81
4 4!6! 4 3 2 1
17,010
x x x
x
⋅ ⋅ ⋅⋅ = ⋅ = ⋅ ⋅ ⋅ ⋅
=
6The coefficient of is 17,010.x
30. 10, 7, , 3n j x x a= = = = −
( )
( )
3 7 3
3
3
10 10!( 3) 2187
7 7!3!
10 9 82187
3 2 1
262,440
x x
x
x
⋅ − = ⋅ −
⋅ ⋅= ⋅ −⋅ ⋅
= −
3The coefficient of is 262,440.x −
31. 12, 5, 2 , 1n j x x a= = = = −
7 5 7
7
7
12 12!(2 ) ( 1) 128 ( 1)
5 5!7!
12 11 10 9 8( 128)
5 4 3 2 1
101,376
x x
x
x
⋅ − = ⋅ −
⋅ ⋅ ⋅ ⋅= ⋅ −⋅ ⋅ ⋅ ⋅
= −
7The coefficient of is 101,376.x −
32. 12, 9, 2 , 1n j x x a= = = =
3 9 3
3
3
12 12!(2 ) (1) 8 (1)
9 9!3!
12 11 108
3 2 1
1760
x x
x
x
⋅ = ⋅
⋅ ⋅= ⋅⋅ ⋅
=
3The coefficient of is 1760.x
33. 9, 2, 2 , 3n j x x a= = = =
7 2 7
7
7
9 9!(2 ) 3 128 (9)
2 2!7!
9 8128 9
2 1
41,472
x x
x
x
⋅ = ⋅
⋅= ⋅ ⋅⋅
=
7The coefficient of is 41,472.x
34. 9, 7, 2 , 3n j x x a= = = = −
2 7 2
2
2
9 9!(2 ) ( 3) 4 ( 2187)
7 7! 2!
9 84 2187
2 1
314,928
x x
x
x
⋅ − = ⋅ −
⋅= ⋅ ⋅ −⋅
= −
2The coefficient of is 314,928.x −
35. 7, 4, , 3n j x x a= = = =
3 4 3 3 37 7! 7 6 53 81 81 2835
4 4!3! 3 2 1x x x x
⋅ ⋅⋅ = ⋅ = ⋅ = ⋅ ⋅
36. 7, 2, , 3n j x x a= = = = −
5 2 5 5 57 7! 7 6( 3) 9 9 189
2 2!5! 2 1x x x x
⋅⋅ − = ⋅ = ⋅ = ⋅
37. 9, 2, 3 , 2n j x x a= = = = −
7 2 7
7 7
9 9!(3 ) ( 2) 2187 4
2 2!7!
9 88748 314,928
2 1
x x
x x
⋅ − = ⋅ ⋅
⋅= ⋅ =⋅
Section 12.5: The Binomial Theorem
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38. 8, 5, 3 , 2n j x x a= = = =
3 5 3
3 3
8 8!(3 ) ( 2) 27 32
5 5!3!
8 7 6864 48,384
3 2 1
x x
x x
⋅ = ⋅ ⋅
⋅ ⋅= ⋅ =⋅ ⋅
39. The 0x term in
( )12 12122 24 3
0 0
12 121j
j j
j j
x xj jx
− −
= =
=
occurs when: 24 3 0
24 3
8
j
j
j
− ===
The coefficient is
12 12! 12 11 10 9495
8! 4! 4 3 2 18⋅ ⋅ ⋅ = = = ⋅ ⋅ ⋅
40. The 0x term in
( ) ( )9 9
9 9 32
0 0
9 911
jj j j
j j
x xj jx
− −
= =
− = −
occurs when: 9 3 0
9 3
3
j
j
j
− ===
The coefficient is
( )39 9! 9 8 71 84
3 3!6! 3 2 1
⋅ ⋅− = − = − = − ⋅ ⋅
41. The 4x term in
( ) ( )310 10 1010 2
0 0
10 1022
jjj j
j j
x xj jx
−−
= =
− = −
occurs when: 3
10 423
62
4
j
j
j
− =
− = −
=
The coefficient is
( )410 10! 10 9 8 72 16 16 3360
4 6! 4! 4 3 2 1
⋅ ⋅ ⋅− = ⋅ = ⋅ = ⋅ ⋅ ⋅
42. The 2x term in
( ) ( )8 88 4
0 0
8 833
jj j j
j j
x xj jx
− −
= =
=
occurs when: 4 2
2
2
j
j
j
− =− = −
=
The coefficient is
( )28 8! 8 73 9 9 252
2 6! 2! 2 1
⋅= ⋅ = ⋅ = ⋅
43. ( ) ( ) ( )5 2 35 3 5 4 3 3 3 2 35 5 5 5(1.001) 1 10 1 1 10 1 10 1 10
0 1 2 3
1 5(0.001) 10(0.000001) 10(0.000000001)
1 0.005 0.000010 0.000000010
1.00501 (correct to 5 decimal places)
− − − − = + = + ⋅ + ⋅ + ⋅ + ⋅⋅⋅
+ + + + ⋅⋅ ⋅
= + + + + ⋅⋅ ⋅=
=
44. ( ) ( ) ( )
( ) ( ) ( )
6 2 36 6 5 4 36 6 6 6(0.998) 1 0.002 1 1 ( 0.002) 1 0.002 1 0.002
0 1 2 3
1 6 0.002 15 0.000004 20 0.000000008 ...
1 0.012 0.00006 0.00000016
0.98806 (correct to 5 decimal places)
= − = + ⋅ − + ⋅ − + ⋅ ⋅ − + ⋅⋅⋅
= + − + + − += − + −=
Chapter 12: Sequences; Induction; the Binomial Theorem
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45. ( ) ( ) ( )
( )( )
1 !! !1 1 !( 1 )! 1 !(1)! 1 !
n nn n nn
n n n n n n
− = = = = − − − − − −
! ! ! !1
!( )! !0! ! 1 !
n n n n n
n n n n n n n
= = = = = − ⋅
46. ( )
! ! !
( )! ( ( ))! ! ! !( )!
n nn n n
n j jn j n n j n j j j n j
= = = = − − − − − −
47. Show that 20 1
nn n n
n
+ + ⋅⋅ ⋅ + =
1 2 2
2 (1 1)
1 1 1 1 1 1 10 1 2
0 1
n n
n n n n n nn n n n
n
n n n
n
− − −
= +
= ⋅ + ⋅ ⋅ + ⋅ ⋅ + ⋅⋅ ⋅ + ⋅ ⋅
= + + ⋅⋅ ⋅ +
48. Show that ( 1) 00 1 2
nn n n n
n
− + − ⋅⋅ ⋅+ − =
1 2 2
0 (1 1)
1 1 ( 1) 1 ( 1) 1 ( 1)0 1 2
( 1)0 1 2
n
n n n n n n
n
n n n n
n
n n n n
n
− − −
= −
= ⋅ + ⋅ ⋅ − + ⋅ ⋅ − + ⋅⋅ ⋅ + ⋅ ⋅ −
= − + − ⋅⋅ ⋅ + −
49. 5 4 3 2 2 3 4 5 5
55 5 5 5 5 51 1 3 1 3 1 3 1 3 3 1 3(1) 1
0 1 2 3 4 54 4 4 4 4 4 4 4 4 4 4 4
+ + + + + = + = =
50. 812! 479,001,600 4.790016 10= = × 18
25
20! 2.432902008 10
25! 1.551121004 10
≈ ×
≈ ×
12
2018
2525
12 112! 2 12 1 479,013,972.4
12 12 1
20 120! 2 20 1 2.43292403 10
12 20 1
25 125! 2 25 1 1.551129917 10
12 25 1
e
e
e
≈ ⋅ π + ≈ ⋅ −
≈ ⋅ π + ≈ × ⋅ −
≈ ⋅ π + ≈ × ⋅ −
Chapter 12 Review Exercises
1287
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Chapter 12 Review Exercises
1. 1 2 3 4 51 2 3 4 5
1 3 4 2 3 5 3 3 6 4 3 7 5 3 8( 1) , ( 1) , ( 1) , ( 1) , ( 1)
1 2 3 2 2 4 3 2 5 4 2 6 5 2 7a a a a a
+ + + + += − = − = − = = − = − = − = = − = −+ + + + +
2. 1 1 2 1 3 11 2 3( 1) (2 1 3) 5, ( 1) (2 2 3) 7, ( 1) (2 3 3) 9,b b b+ + += − ⋅ + = = − ⋅ + = − = − ⋅ + =
4 1 5 14 5( 1) (2 4 3) 11, ( 1) (2 5 3) 13b b+ += − ⋅ + = − = − ⋅ + =
3. 1 2 3 4 5
1 2 3 4 52 2 2 2 2
2 2 2 4 2 8 2 16 2 322, 1, , 1,
1 4 9 16 251 2 3 4 5c c c c c= = = = = = = = = = = = =
4. 1 2 3 4 5
1 2 3 4 5, , , ,1 2 3 4 5
e e e e ed e d d d d= = = = = =
5. 1 2 3 4 52 2 4 2 4 8 2 8 16
3, 3 2, 2 , ,3 3 3 3 3 9 3 9 27
a a a a a= = ⋅ = = ⋅ = = ⋅ = = ⋅ =
6. 1 2 3 4 51 1 1 1 1 1 1 1 1
4, 4 1, 1 , ,4 4 4 4 4 16 4 16 64
a a a a a= = − ⋅ = − = − ⋅ − = = − ⋅ = − = − ⋅ − =
7. 1 2 3 4 52, 2 2 0, 2 0 2, 2 2 0, 2 0 2a a a a a= = − = = − = = − = = − =
8. 1 2 3 4 53, 4 ( 3) 1, 4 1 5, 4 5 9, 4 9 13a a a a a= − = + − = = + = = + = = + =
9. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )4
1
(4 2) 4 1 2 4 2 2 4 3 2 4 4 2 6 10 14 18 48k
k=
+ = ⋅ + + ⋅ + + ⋅ + + ⋅ + = + + + =
10. ( ) ( ) ( ) ( ) ( ) ( )3
2 2 2 2
1
(3 ) 3 1 3 2 3 3 2 1 6 5k
k=
− = − + − + − = + − + − = −
11. ( )13
1
1
1 1 1 1 11 1
2 3 4 13k
k k+
=
− + − + ⋅⋅ ⋅+ = −
12. 2 3 4 1 1
2 30
1
11
2 2 2 2 22
3 3 3 3 3
2
3
n kn
n kk
kn
kk
+ +
=
+
−=
+ + + + ⋅⋅ ⋅+ =
=
13. { } { }5na n= + Arithmetic
[ ]
( 1 5) ( 5) 6 5 1
6 5 ( 11)2 2n
d n n n n
n nS n n
= + + − + = + − − =
= + + = +
14. { } { }4 3nb n= + Arithmetic
[ ]
( )
(4( 1) 3) (4 3)
4 4 3 4 3 4
7 4 32
(4 10)2
2 5
n
d n n
n n
nS n
nn
n n
= + + − += + + − − =
= + +
= +
= +
15. { } { }32nc n= Examine the terms of the
sequence: 2, 16, 54, 128, 250, ... There is no common difference; there is no common ratio; neither.
16. { } { }22 1nd n= − Examine the terms of the
sequence: 1, 7, 17, 31, 49, ...
Chapter 12: Sequences; Induction; the Binomial Theorem
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There is no common difference; there is no common ratio; neither.
17. { } { }32 nns = Geometric
3( 1) 3 3
3 3 3 33 3
2 22 2 8
2 2
n nn n
n nr
+ ++ −= = = = =
( )1 8 1 8 88 8 8 1
1 8 7 7
n nn
nS − −= = = − − −
18. { } { }23 nnu = Geometric
2( 1) 2 2
2 2 2 22 2
3 33 3 9
3 3
n nn n
n nr
+ ++ −= = = = =
( )1 9 1 9 99 9 9 1
1 9 8 8
n nn
nS − −= = = − − −
19. 0, 4, 8, 12, ... Arithmetic 4 0 4d = − =
( ) ( )2(0) ( 1)4 4( 1) 2 ( 1)2 2nn n
S n n n n= + − = − = −
20. 1, –3, –7, –11, ... Arithmetic 3 1 4d = − − = −
( ) ( )
( )
2(1) ( 1)( 4) 2 4 42 2
(6 4 ) 3 22
nn n
S n n
nn n n
= + − − = − +
= − = −
21. 3 3 3 3
3, , , , , ...2 4 8 16
Geometric
33 1 12
3 2 3 2r
= = ⋅ =
1 11 1
12 23 3 6 1
1 1 212 2
n n
n
nS
− − = = = − −
22. 5 5 5 5
5, , , , , ...3 9 27 81
− − Geometric
55 1 13
5 3 5 3r
− = = − ⋅ = −
1 11 1
3 35 5
1 41
3 3
15 11
4 3
n n
n
n
S
− − − − = = − −
= − −
23. Neither. There is no common difference or common ratio.
24. 3 5 7 9 11
, , , , ,2 4 6 8 10
Neither. There is no
common difference or common ratio.
25. ( )50 50
1 1
50 50 13 3 3 3825
2k k
k k= =
+ = = =
26. ( )( )30
2
1
30 30 1 2 30 19455
6k
k=
+ ⋅ += =
27. 30 30 30 30 30
1 1 1 1 1
(3 9) 3 9 3 9
30(30 1)3 30(9)
2
1395 270 1125
k k k k k
k k k= = = = =
− = − = −
+ = −
= − =
28. 40 40 40
1 1 1
40 40
1 1
( 2 8) 2 8
2 8
40(1 40)2 40(8)
2
1640 320 1320
k k k
k k
k k
k
= = =
= =
− + = − +
= − +
+ = − +
= − + = −
Chapter 12 Review Exercises
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29.
7 7
7
1
1 11 1
1 1 13 31 23 3 313 3
1 11
2 2187
1 2186 10930.49977
2 2187 2187
k
k =
− − = = − = −
= ⋅ = ≈
30. ( ) ( )
( )
1010
1
1 22 2
1 ( 2)
1 1024 22 1023
3 3
682
k
k =
− − − = − − −
− = − = − −
=
31. Arithmetic
1 13, 4, ( 1)na d a a n d= = = + −
9 3 (9 1)4 3 8(4) 3 32 35a = + − = + = + =
32. Arithmetic
1 11, 2, ( 1)na d a a n d= = − = + − 8 1 (8 1)( 2) 1 7( 2) 1 14 13a + − − = + − = − = −=
33. Geometric
11 1
11, , 11;
10n
na r n a a r −= = = =
11 1 10
111 1
110 10
1
10,000,000,000
a−
= ⋅ =
=
34. Geometric 1
1 11, 2, 11; nna r n a a r −= = = =
( ) ( )11 1 1011 1 2 2 1024a
−= ⋅ = =
35. Arithmetic
1 12, 2, 9, ( 1)na d n a a n d= = = = + −
9 2 (9 1) 2 2 8 2
9 2 12.7279
a = + − = +
= ≈
36. Geometric
11 12, 2, 9, n
na r n a a r −= = = =
( ) ( )9 1 8
9 2 2 2 2 2 16
16 2 22.6274
a−
= = = ⋅
= ≈
37. 7 1 20 16 31 19 96a a d a a d= + = = + = ;
Solve the system of equations:
1
1
6 31
19 96
a d
a d
+ =+ =
Subtract the second equation from the first equation and solve for d.
13 65
5
d
d
− = −=
1 31 6(5) 31 30 1a = − = − =
( )( )( )
1 1
1 1 5
1 5 5
5 4
na a n d
n
n
n
= + −
= + −= + −= −
General formula: { } { }5 4na n= −
38. 8 1 17 17 20 16 47a a d a a d= + = − = + = − ;
Solve the system of equations:
1
1
7 20
16 47
a d
a d
+ = −+ = −
Subtract the second equation from the first equation and solve for d.
9 27
3
d
d
− == −
1 20 7( 3) 20 21 1a = − − − = − + =
( )( )( )
1 1
1 1 3
1 3 3
3 4
na a n d
n
n
n
= + −
= + − −= − += − +
General formula: { } { }3 4na n= − +
39. 10 1 18 19 0 17 8a a d a a d= + = = + = ;
Solve the system of equations:
1
1
9 0
17 8
a d
a d
+ =+ =
Subtract the second equation from the first equation and solve for d.
8 8
1
d
d
− = −=
1 9(1) 9a = − = −
Chapter 12: Sequences; Induction; the Binomial Theorem
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( )( )( )
1 1
9 1 1
9 1
10
na a n d
n
n
n
= + −
= − + −= − + −= −
General formula: { } { }10na n= −
40. 12 1 22 111 30 21 50a a d a a d= + = = + = ;
Solve the system of equations:
1
1
11 30
21 50
a d
a d
+ =+ =
Subtract the second equation from the first equation and solve for d.
10 20
2
d
d
− = −=
1 30 11(2) 30 22 8a = − = − =
( )( ) ( )
1 1
8 1 2
8 2 2
2 6
na a n d
n
n
n
= + −
= + −= + −= +
General formula: { } { }2 6na n= +
41. 11
3,3
a r= =
1
Since 1, the series converges.
3 3 91 21 2
13 3
n
r
aS
r
<
= = = =− −
42. 11
2,2
a r= =
1
Since 1, the series converges.
2 24
1 111
2 2
n
r
aS
r
<
= = = =− −
43. 11
2,2
a r= = −
Since 1r < , the series converges.
1 2 2 431 31
122
na
Sr
= = = =− − −
44. 12
6,3
a r= = −
Since 1r < , the series converges.
1 6 6 1851 52
133
na
Sr
= = = =− − −
45. 11
2a = ,
3
2r =
Since 1r > , the series diverges.
46. 1 5a = , 5
4r = −
Since 1r > , the series diverges.
47. 11
4,2
a r= =
Since 1r < , the series converges.
1 4 48
1 111
2 2
na
Sr
= = = =− −
48. 13
3,4
a r= = −
Since 1r < , the series converges.
1 3 3 1271 73
144
na
Sr
= = = =− − −
49. I: 3 1
1: 3 1 3 and (1 1) 32
n⋅= ⋅ = + =
II: If 3
3 6 9 3 ( 1)2
kk k+ + + + = + , then
[ ]3 6 9 3 3( 1)
3 6 9 3 3( 1)
3( 1) 3( 1)
23 3( 1)
( 1) 3 (( 1) 1)2 2
k k
k k
kk k
k kk k
+ + + + + += + + + + + +
= + + +
+ = + + = + +
Conditions I and II are satisfied; the statement is true.
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50. I: 21: 4 1 2 2 and 2(1) 2n = ⋅ − = =
II: If 22 6 10 (4 2) 2k k+ + + + − = , then
[ ]
( ) ( )
2
22
2 6 10 (4 2) (4( 1) 2)
2 6 10 (4 2) 4 2
2 4 2
2 2 1 2 1
k k
k k
k k
k k k
+ + + + − + + −= + + + + − + +
= + +
= + + = +
Conditions I and II are satisfied; the statement is true.
51. I: 1 1 11: 2 3 2 and 3 1 2n −= ⋅ = − =
II: If 12 6 18 2 3 3 1k k−+ + + + ⋅ = − , then 1 1 1
1
1
2 6 18 2 3 2 3
2 6 18 2 3 2 3
3 1 2 3 3 3 1 3 1
k k
k k
k k k k
− + −
−
+
+ + + + ⋅ + ⋅
= + + + + ⋅ + ⋅
= − + ⋅ = ⋅ − = −
Conditions I and II are satisfied; the statement is true.
52. I: ( )1 1 11: 3 2 3 and 3 2 1 3n −= ⋅ = − =
II: If ( )13 6 12 3 2 3 2 1k k−+ + + + ⋅ = − , then
( )( ) ( ) ( )
1 1 1
1
1
3 6 12 3 2 3 2
3 6 12 3 2 3 2
3 2 1 3 2
3 2 1 2 3 2 2 1 3 2 1
k k
k k
k k
k k k k
− + −
−
+
+ + + + ⋅ + ⋅
= + + + + ⋅ + ⋅
= − + ⋅
= ⋅ − + = ⋅ − = −
Conditions I and II are satisfied; the statement is true.
53. I: 1:n =
2 21(3 1 2) 1 and 1(6 1 3 1 1) 1
2⋅ − = ⋅ ⋅ − ⋅ − =
II: If
( )2 2 2 211 4 (3 2) 6 3 1
2k k k k+ + + − = ⋅ − − ,
then
( )
( )
( ) ( ) ( )
22 2 2 2
2 2 2 2 2
2 2
3 2 2
3 2
2
3 2 2
2
1 4 7 (3 2) 3( 1) 2
1 4 7 (3 2) (3 1)
16 3 1 (3 1)
21
6 3 18 12 221
6 15 11 221
( 1) 6 9 221
6 6 9 9 2 221
6 1 9 1 2 12
k k
k k
k k k k
k k k k k
k k k
k k k
k k k k k
k k k k k
+ + + + − + + −
= + + + + − + +
= ⋅ − − + +
= ⋅ − − + + +
= ⋅ + + +
= ⋅ + + +
= ⋅ + + + + +
= ⋅ + + + + +
2
2
2
1( 1) 6 12 6 3 3 1
21
( 1) 6( 2 1) 3( 1) 121
( 1) 6( 1) 3( 1) 12
k k k k
k k k k
k k k
= ⋅ + + + − − −
= ⋅ + + + − + −
= ⋅ + + − + −
Conditions I and II are satisfied; the statement is true.
Chapter 12: Sequences; Induction; the Binomial Theorem
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54. I: 1
1: 1(1 2) 3 and (1 1)(2 1 7) 36
n = + = ⋅ + ⋅ + =
II: If
1 3 2 4 ( 2) ( 1)(2 7)6
kk k k k⋅ + ⋅ + + + = + + ,
then
[ ]
( )( )( )
2
2
1 3 2 4 ( 2) ( 1)( 1 2)
1 3 2 4 ( 2) ( 1)( 3)
( 1)(2 7) ( 1)( 3)6( 1)
2 7 6 186
( 1)2 13 18
6( 1)
( 1) 1 (2( 1) 7)6
k k k k
k k k k
kk k k k
kk k k
kk k
kk k
⋅ + ⋅ + + + + + + += ⋅ + ⋅ + + + + + +
= + + + + +
+= + + +
+= + +
+= + + + +
Conditions I and II are satisfied; the statement is true.
55. 5 5! 5 4 3 2 1 5 4
102 2!3! 2 1 3 2 1 2 1
⋅ ⋅ ⋅ ⋅ ⋅= = = = ⋅ ⋅ ⋅ ⋅ ⋅
56. 8 8! 8 7 6 5 4 3 2 1 8 7
286 6! 2! 6 5 4 3 2 1 2 1 2 1
⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅= = = = ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅
57. 5 5 4 3 2 2 3 1 4 5
5 4 3 2
5 4 3 2
5 5 5 5 5 5( 2) 2 2 2 2 2
0 1 2 3 4 5
5 2 10 4 10 8 5 16 1 32
10 40 80 80 32
x x x x x x
x x x x x
x x x x x
+ = + ⋅ + ⋅ + ⋅ + ⋅ + ⋅
= + ⋅ + ⋅ + ⋅ + ⋅ + ⋅
= + + + + +
58. 4 4 3 2 2 3 0 4
4 3 2
4 3 2
4 4 4 4 4( 3) ( 3) ( 3) ( 3) ( 3)
0 1 2 3 4
4( 3) 6 9 4( 27) 81
12 54 108 81
x x x x x x
x x x x
x x x x
− = + − + − + − + −
= + − + ⋅ + − +
= − + − +
59. 5 5 4 3 2 2 3 1 4 5
5 4 3 2
5 4 3 2
5 5 5 5 5 5(2 3) (2 ) (2 ) 3 (2 ) 3 (2 ) 3 (2 ) 3 3
0 1 2 3 4 5
32 5 16 3 10 8 9 10 4 27 5 2 81 1 243
32 240 720 1080 810 243
x x x x x x
x x x x x
x x x x x
+ = + ⋅ + ⋅ + ⋅ + ⋅ + ⋅
= + ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ + ⋅
= + + + + +
60. 4 4 3 2 2 3 4
4 3 2
4 3 2
4 4 4 4 4(3 4) (3 ) (3 ) ( 4) (3 ) ( 4) (3 )( 4) ( 4)
0 1 2 3 4
81 4 27 ( 4) 6 9 16 4 3 ( 64) 1 256
81 432 864 768 256
x x x x x
x x x x
x x x x
− = + − + − + − + −
= + ⋅ − + ⋅ ⋅ + ⋅ − + ⋅
= − + − +
61. 9, 2, , 2n j x x a= = = =
7 2 7 7 7
7
9 9! 9 82 4 4 144
2 2!7! 2 1
The coefficient of is 144.
x x x x
x
⋅⋅ = ⋅ = ⋅ = ⋅
Chapter 12 Review Exercises
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62. 8, 5, , 3n j x x a= = = = −
3 5 3 3
3
8 8! 8 7 6( 3) ( 243) ( 243)
5 5!3! 3 2 1
13,608
x x x
x
⋅ ⋅− = − = − ⋅ ⋅
= −
3The coefficient of is 13,608.x −
63. 7, 5, 2 , 1n j x x a= = = =
2 5 2 2 2
2
7 7! 7 6(2 ) 1 4 (1) 4 84
5 5! 2! 2 1
The coefficient of is 84.
x x x x
x
⋅⋅ = ⋅ = ⋅ = ⋅
64. 8, 2, 2 , 1n j x x a= = = =
6 2 6
6 6
8 8!(2 ) 1 64 (1)
2 2!6!
8 764 1792
2 1
x x
x x
⋅ = ⋅
⋅= ⋅ =⋅
6The coefficient of is 1792.x
65. This is an arithmetic sequence with
1 80, 3, 25a d n= = − =
a. 25 80 (25 1)( 3) 80 72 8 bricksa = + − − = − =
b. 2525
(80 8) 25(44) 1100 bricks2
S = + = =
1100 bricks are needed to build the steps.
66. This is an arithmetic sequence with
1 30, 1, 15na d a= = − =
15 30 ( 1)( 1)
15 1
16
16
n
n
n
n
= + − −− = − +− = −
=
1616
(30 15) 8(45) 360 tiles2
S = + = =
360 tiles are required to make the trapezoid.
67. This is a geometric sequence with
13
20,4
a r= = .
a. After striking the ground the third time, the
height is 3
3 13520 8.44 feet
4 16 = ≈
.
b. After striking the ground the thn time, the
height is 3
20 feet4
n
.
c. If the height is less than 6 inches or 0.5 feet, then:
( )
( )
30.5 20
4
30.025
4
3log 0.025 log
4
log 0.02512.82
3log
4
n
n
n
n
≥
≥
≥
≥ ≈
The height is less than 6 inches after the 13th strike.
d. Since this is a geometric sequence with 1r < , the distance is the sum of the two
infinite geometric series - the distances going down plus the distances going up. Distance going down:
20 2080
3 11
4 4
downS = = = −
feet.
Distance going up: 15 15
603 1
14 4
upS = = = −
feet.
The total distance traveled is 140 feet.
68. This is an ordinary annuity with $200P = and
( ) ( )12 20 240n = = payment periods. The
Chapter 12: Sequences; Induction; the Binomial Theorem
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interest rate per period is 0.10
12. Thus,
2400.10
1 112
200 $151,873.770.1012
A
+ − = ≈
69. This is an ordinary annuity with $500P = and
( )( )4 30 120n = = payment periods. The
interest rate per period is 0.08
0.024
= . Thus,
[ ]1201 0.02 1
500 $244,129.080.02
A + − = ≈
70. This is a geometric sequence with
1 20,000, 1.04, 5a r n= = = . Find the fifth
term of the sequence: 5 1 4
5 20,000(1.04) 20,000(1.04)
23,397.17
a −= ==
Her salary in the fifth year will be $23,397.17.
Chapter 12 Test
1. 2 1
8nn
an
−=+
2
1
2
2
2
3
2
4
2
5
1 1 00
1 8 9
2 1 32 8 10
3 1 83 8 11
4 1 15 54 8 12 4
5 1 245 8 13
a
a
a
a
a
−= = =+−= =
+−= =
+−= = =
+−= =
+
The first five terms of the sequence are 0, 3
10,
811
, 54
, and 2413
.
2. 1 14; 3 2n na a a −= = +
( )( )( )( )
2 1
3 2
4 3
5 4
3 2 3 4 2 14
3 2 3 14 2 44
3 2 3 44 2 134
3 2 3 134 2 404
a a
a a
a a
a a
= + = + =
= + = + =
= + = + =
= + = + =
The first five terms of the sequence are 4, 14, 44, 134, and 404.
3. ( )
( ) ( ) ( )
( ) ( ) ( )
2 2 2
31
21
1 1 2 1 3 1
2 3 4
1 1 2 1 3 1
1 2 3
11
1 1 1
2 3 41 1 1
1 4 9
3 4 612
4 9 36
k
k
k
k
+
=
+ + ++ + +
+ −
= − + − + − = − + − + −
= − + =
4.
( ) ( ) ( ) ( )1 2 3 4
4
1
2 2 2 21 2 3 4
3 3 3 3
23
2 4 8 161 2 3 4
3 9 27 81130 680
1081 81
k
k
k=
− + − + − + −=
=
−
− + − + − + −
= − = −
5. 2 3 4 11
...5 6 7 14
− + − + +
Notice that the signs of each term alternate, with the first term being negative. This implies that the general term will include a power of 1− . Also note that the numerator is always 1 more than the term number and the denominator is 4 more than the term number. Thus, each term is in
the form ( ) 11
4k k
k+ − +
. The last numerator is 11
which indicates that there are 10 terms.
( )10
1
2 3 4 11 1... 1
5 6 7 14 4k
k
kk=
+ − + − + + = − +
6. 6,12,36,144,... 12 6 6− = and 36 12 24− = The difference between consecutive terms is not constant. Therefore, the sequence is not arithmetic. 126 2= and 36
12 3=
Chapter 12 Test
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The ratio of consecutive terms is not constant. Therefore, the sequence is not geometric.
7. 1
42
nna = − ⋅
11 12 2
1 11 11 2 2
4 4 44
4 4
n nn
n nn
aa
−
− −−
− ⋅ − ⋅ ⋅= = =
− ⋅ − ⋅
Since the ratio of consecutive terms is constant, the sequence is geometric with common ratio
4r = and first term 11
14 2
2a = − ⋅ = − .
The sum of the first n terms of the sequence is given by
( )
111
1 42
1 42
1 43
n
n
n
n
rS a
r−= ⋅−−= − ⋅−
= −
8. 2, 10, 18, 26,...− − − −
( )10 2 8− − − = − , ( )18 10 8− − − = − ,
( )26 18 8− − − = −
The difference between consecutive terms is constant. Therefore, the sequence is arithmetic with common difference 8d = − and first term
1 2a = − .
( )( ) ( )
1 1
2 1 8
2 8 8
6 8
na a n d
n
n
n
= + −
= − + − −= − − += −
The sum of the first n terms of the sequence is given by
( )
( )
( )( )
2
2 6 82
4 82
2 4
n nn
S a a
nn
nn
n n
= +
= − + −
= −
= −
9. 72nn
a = − +
( )1
17 7
2 2
17 7
2 212
n n
nna a
n n
−− − = − + − − +
−= − + + −
= −
The difference between consecutive terms is constant. Therefore, the sequence is arithmetic
with common difference 12
d = − and first term
11 13
72 2
a = − + = .
The sum of the first n terms of the sequence is given by
( )
( )
12
137
2 2 2
272 2 2
274
n nn
S a a
n n
n n
nn
= +
= + − + = −
= −
10. 8
25,10, 4, ,...5
10 225 5
= , 4 2
10 5= ,
85 8 1 24 5 4 5
= ⋅ =
The ratio of consecutive terms is constant. Therefore, the sequence is geometric with
common ratio 25r = and first term 1 25a = .
The sum of the first n terms of the sequence is given by
1
22 11 55125 25
2 31 15 5
5 2 125 225 1 1
3 5 3 5
nn
n
n
n n
rS a
r
− − − = ⋅ = ⋅ = ⋅− −
= ⋅ − = −
Chapter 12: Sequences; Induction; the Binomial Theorem
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11. 2 32 1nn
an
−=+
( )( )
( )( ) ( )( )( )( )
( ) ( )
1
2 2
2
2
2 1 32 3 2 3 2 52 1 2 1 2 12 1 1
2 3 2 1 2 5 2 1
2 1 2 1
4 8 3 4 8 5
4 18
4 1
n n
nn n na a
n n nn
n n n n
n n
n n n n
n
n
−− −− − −− = − = −
+ + −− +
− − − − +=
+ −
− + − − −=
−
=−
The difference of consecutive terms is not constant. Therefore, the sequence is not arithmetic.
( )( )
( ) ( )( )( )1
2 32 3 2 12 3 2 12 1
2 1 2 52 1 3 2 1 2 5
2 1 1
n
n
nn na n nn
a n nn n n
n−
−− −− −+= = ⋅ =
+ −− − + −− +
The ratio of consecutive terms is not constant. Therefore, the sequence is not geometric.
12. For this geometric series we have 64 1
256 4r
−= = −
and 1 256a = . Since 1 1
14 4
r = − = < , the series
converges and we get
( )1
5144
256 256 10241 51
aS
r∞ = = = =− − −
13. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( )5 5 4 3 2 2 3 4 5
5 4 3 2
5 4 3 2
5 5 5 5 5 53 2 3 3 2 3 2 3 2 3 2 2
0 1 2 3 4 5
243 5 81 2 10 27 4 10 9 8 5 3 16 32
243 810 1080 720 240 32
m m m m m m
m m m m m
m m m m m
+ = + + + + +
= + ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ +
= + + + + +
14. First we show that the statement holds for 1n = .
11 1 1 2
1 + = + =
The equality is true for 1n = so Condition I holds. Next we assume that 1 1 1 1
1 1 1 ... 1 11 2 3
nn
+ + + + = +
is
true for some k, and we determine whether the formula then holds for 1k + . We assume that
1 1 1 11 1 1 ... 1 1
1 2 3k
k + + + + = +
.
Now we need to show that ( )1 1 1 1 11 1 1 ... 1 1 1 1 2
1 2 3 1k k
k k + + + + + = + + = + +
.
We do this as follows:
( )
( ) ( )
1 1 1 1 1 1 1 1 1 11 1 1 ... 1 1 1 1 1 ... 1 1
1 2 3 1 1 2 3 1
11 1 (using the induction assumption)
1
11 1 1 1 1
12
k k k k
kk
k k kk
k
+ + + + + = + + + + + + + = + + +
= + ⋅ + + ⋅ = + ++
= +
Condition II also holds. Thus, formula holds true for all natural numbers.
15. The yearly values of the car form a geometric sequence with first term 1 31,000a = and
common ratio 0.85r = (which represents a 15% loss in value).
Chapter 12 Cumulative Review
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( ) 131,000 0.85
nna
−= ⋅
The nth term of the sequence represents the value of the car at the beginning of the nth year. Since we want to know the value after 10 years, we are looking for the 11th term of the sequence. That is, the value of the car at the beginning of the 11th year.
( )1011 111 1 31,000 0.85 6,103.11a a r −= ⋅ = ⋅ =
After 10 years, the car will be worth $6,103.11.
16. The weights for each set form an arithmetic sequence with first term 1 100a = and common
difference 30d = . If we imagine the weightlifter only performed one repetition per set, the total weight lifted in 5 sets would be the sum of the first five terms of the sequence.
( )( )( ) ( )
1
5
1
100 5 1 30 100 4 30 220
na a n d
a
= + −
= + − = + =
( )
( ) ( )55 52 2
2
100 220 320 800
n nn
S a a
S
= +
= + = =
Since he performs 10 repetitions in each set, we multiply the sum by 10 to obtain the total weight lifted.
( )10 800 8000=
The weightlifter will have lifted a total of 8000 pounds after 5 sets.
Chapter 12 Cumulative Review
1. 2 9x =
2 29 or 9
3 or 3
x x
x x i
= = −= ± = ±
2. a. graphing 2 2 100x y+ = and 23y x= .
b. solving 32 2 2 2
2 2
2
100 3 3 300
3 3 0
3 300
x y x y
y x x y
y y
+ = ⎯⎯→ + =
= ⎯⎯→ − + =
+ =
2 23 300 3 300 0y y y y+ = + − =
( )( )( )
21 1 4 3 300 1 3601
2 3 6y
− ± − − − ±= =
Substitute and solve for x:
2
2
1 3601 1 36013
6 6
1 3601 1 3601
18 18
y x
x x
− + − += =
− + − += = ±
or
2
2
1 3601 1 36013
6 6
1 3601 1 3601
18 18
1 3601undefined since 0
18
y x
x x
− − − −= =
− − − −= = ±
− − <
Therefore, the system has solutions
1 3601 1 3601,
18 6x y
− + − += = and
1 3601 1 3601,
18 6x y
− + − += − = .
1 3601 1 3601, ,
18 6
− + − +
1 3601 1 3601,
18 6
− + − +−
c. The graphs intersect at the points
( )1 3601 1 3601, 1.81,9.84
18 6
− + − + ≈
( )1 3601 1 3601,
18 61.81,9.84
− + − +− ≈ −
Chapter 12: Sequences; Induction; the Binomial Theorem
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3. 2 5xe =
( )
5
25
ln ln2
5ln 0.916
2
x
x
e
e
x
=
=
= ≈
The solution set is 5
ln2
.
4. slope 5m= = ; Since the x-intercept is 2, we
know the point ( )2,0 is on the graph of the line
and is a solution to the equation 5y x b= + .
( )5
0 5 2
0 10
10
y x b
b
b
b
= += += +
− =
Therefore, the equation of the line with slope 5 and x-intercept 2 is 5 10y x= − .
5. Given a circle with center (–1, 2) and containing the point (3, 5), we first use the distance formula to determine the radius.
( )( ) ( )2 2
2 2
3 1 5 2
4 3 16 9
25
5
r = − − + −
= + = +
==
Therefore, the equation of the circle is given by
( )( ) ( )( ) ( )
2 2 2
2 2 2
2 2
2 2
1 2 5
1 2 5
2 1 4 4 25
2 4 20 0
x y
x y
x x y y
x y x y
− − + − =
+ + − =
+ + + − + =
+ + − − =
6. ( ) 3
2
xf x
x=
−, ( ) 2 1g x x= +
a. ( ) ( )
( ) ( )2 2 2 1 5
3 5 155 5
5 2 3
g
f
= + =
= = =−
( )( ) ( )( ) ( )2 2 5 5f g f g f= = =
b. ( ) ( )
( ) ( )
3 4 124 6
4 2 26 2 6 1 13
f
g
= = =−
= + =
( )( ) ( )( ) ( )4 4 6 13g f g f g= = =
c. ( )( ) ( )( )( )
( )3 2 1
2 1 2
6 3
2 1
f g x f g x
x
x
x
x
=
+=
+ −+=−
d. To determine the domain of the composition
( )( )f g x , we start with the domain of g
and exclude any values in the domain of g that make the composition undefined.
( )g x is defined for all real numbers and
( )( )f g x is defined for all real numbers
except 1
2x = . Therefore, the domain of the
composite ( )( )f g x is 1
|2
x x ≠
.
e. ( )( ) 32 1
2
61
26 2
27 2
2
xg f x
x
x
xx x
xx
x
= + −
= +−
+ −=−
−=−
f. To determine the domain of the composition
( )( )g f x , we start with the domain of f
and exclude any values in the domain of f that make the composition undefined.
( )f x is defined for all real numbers except
2x = and ( )( )g f x is defined for all real
numbers except 2x = . Therefore, the
domain of the composite ( )( )g f x is
{ }| 2x x ≠ .
Chapter 12 Cumulative Review
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g. ( ) 2 1g x x= +
2 1
2 1
1 2
1
2
y x
x y
x y
xy
= += +
− =− =
( )1 1
2
xg x− −=
The domain of ( )1g x− is the set of all real
numbers.
h. ( ) 3
2
xf x
x=
−
( )
( )
3
23
2
2 3
2 3
3 2
3 2
2
3
xy
xy
xy
x y y
xy x y
xy y x
y x x
xy
x
=−
=−
− =− =− =− =
=−
( )1 2
3
xf x
x− =
−
The domain of ( )1f x− is { }| 3x x ≠ .
7. Center: (0, 0); Focus: (0, 3); Vertex: (0, 4); Major axis is the y-axis; 4; 3a c= = .
Find b: 2 2 2 16 9 7 7b a c b= − = − = = Write the equation using rectangular coordinates:
2 2
17 16
x y+ =
8. The focus is ( )1,3− and the vertex is ( )1,2− .
Both lie on the vertical line 1x = − . We have a = 1 since the distance from the vertex to the focus is 1 unit, and since ( )1,3− is above
( )1,2− , the parabola opens up. The equation of
the parabola is:
( ) ( )( )( ) ( )
( ) ( )
2
2
2
4
1 4 1 2
1 4 2
x h a y k
x y
x y
− = −
− − = ⋅ ⋅ −
+ = −
9. Center point (0, 4); passing through the pole (0,4) implies that the radius = 4 using rectangular coordinates:
( ) ( )( ) ( )
2 2 2
2 2 2
2 2
2 2
0 4 4
8 16 16
8 0
x h y k r
x y
x y y
x y y
− + − =
− + − =
+ − + =
+ − =
converting to polar coordinates: 2
2
8 sin 0
8 sin
8sin
r r
r r
r
θθ
θ
− =
==
10. 22sin sin 3 0, 0 2x x x π− − = ≤ ≤
( )( )2sin 3 sin 1 0x x− + =
32sin 3 0 sin , which is impossible
23
sin 1 0 sin 12
x x
x x xπ
− = =
+ = = − =
Solution set 3
2
π
.
11. ( )1cos 0.5− −
We are finding the angle , ,θ π θ π− ≤ ≤ whose
cosine equals 0.5− .
( )1
cos 0.5
2 2cos 0.5
3 3
θ π θ π
θ −
= − − ≤ ≤π π= − =
12. 1
sin , is in Quadrant II4
θ θ=
a. is in Quadrant II cos 0θ θ < 2
2 1cos 1 sin 1
4
1 15 151
16 16 4
θ θ = − − = − −
= − − = − = −
b.
1sin 1 44
tancos 4 1515
4
1 15 15
1515 15
θθθ
= = = − −
= − ⋅ = −
Chapter 12: Sequences; Induction; the Binomial Theorem
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c. sin(2 ) 2sin cos
1 152
4 4
15
8
θ θ θ=
= −
= −
d. 2 2
2 2
cos(2 ) cos sin
15 1
4 4
15 1 14 7
16 16 16 8
θ θ θ= −
= − −
= − = =
e. 2 4 2 2
π θ πθ ππ < < < <
is in Quadrant I2
θ sin 0
2
θ >
151
41 cossin
2 2 2
4 154 4 15
2 8
4 15
2 2
θ θ
− − − = =
+ + = =
+=
Chapter 12 Projects
Project I – Internet Based Project
Answers will vary based on the year that is used. Data used in these solutions will be from 2008.
1. I = net immigration = 887,168 Population for 2008 = 303,824,640
2. r = 0.01416 – 0.00826 = 0.0059
3. 1
1
0
(1 0.0059) 887168
(1.0059) 887168
303824640
n n
n n
p p
p p
p
−
−
= + += +=
4. 1 0
1
1
(1.0059) 887168
(1.0059)(303824640) 887168
306,504,373
p p
p
p
= += +=
The population is predicted to be 306,504,373 in 2009.
5. Actual population in 2009: 307,212,123. The formula’s prediction was lower but fairly close.
6. Birth rate: 48.12 per 1000 population (0.04812) Death rate: 12.64 per 1000 population (0.01264) Population for 2008: 31,367,972 I = net immigration = 6587−
0.04812 0.01264 0.03548r = − =
1
1
0
(1 0.03548) 6587
(1.03548) 6587
31,367,972
n n
n n
p p
p p
p
−
−
= + −= −=
1 0
1
1
(1.03548) 6587
(1.03548)(31367972) 6587
32,474,321
p p
p
p
= −= −=
The population is predicted to be 32,474,321 in 2009.
Actual population in 2009: 32,369,558. The formula’s prediction was higher but fairly close.
7. Answers will vary. This appears to support the article. The growth rate for the U.S. is much smaller than the growth rate for Uganda.
8. It could be but one must consider trends in each of the pieces of data to find if the growth rate is increasing or decreasing over time. The same thing must be examined with respect to the net immigration.
Project II
1. 2, 4, 8, 16, 32, 64
2. length n 2n levels
This is a geometric sequence: 2nna =
Recursive expression: 1 02 , 1n na a a−= =
3. 8
256 2
2 2
8
n
n
n
=
==
Chapter 12 Projects
1301
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Project III
1. 1
0
1 1
1 0
3 2 , 18 3
2, 2, 3, 18
3 3
3 18 2 21 2
3 32
7 , 23
st t dt t
t tt
t t
Q P Q P
P b d c
a a
P PP
P P P
−
− −
−
= − + = −= = = =
− = − → =
+ − −= =
= − =
2.
−10
7
10
−9
tP
1tP−
3. 1
1 1
1 1
17
317
3 2(2) 18 33
1 1
s d
s d
P
Q Q
Q Q
=
= − + = −
= =
2
2 2
2 2
29
917 29
3 2 18 33 9
25 25
3 3
s d
s d
P
Q Q
Q Q
=
= − + = −
= =
The market (supply and demand) are getting closer to being the same.
4. The equilibrium price is 4.20.
5. It takes 17 time periods.
6. 17
17
18 3(4.20) 5.40
3 2(4.20) 5.40d
s
Q
Q
= − == − + =
The equilibrium quantity is 5.4.
Project IV
1. 1, 2, 4, 7, 11, 16, 22, 29
2. It is not arithmetic because there is no common difference. It is not geometric because there is no common ration.
3. Scatter diagram 15
−2
6−2
4. 2.5 2.5y x= − The graph does not pass through any of the points.
6
7
8
12.5
15
17.5
y
y
y
===
1
2
3
4
5
0
2.5
5
7.5
10
y
y
y
y
y
=====
5
1( )
(0 1) (2.5 2) (5 4) (7.5 7) (10 11)
1
ir ii
y y=
−
= − + − + − + − + −= −
This is the sum of the errors.
5. 20.5 0.5 1y x x= − +
The graph passes through all of the points.
6
7
8
16
22
29
y
y
y
===
1
2
3
4
5
1
2
4
7
11
y
y
y
y
y
=====
5
1
( ) 0ir i
i
y y=
− =
The sum of the errors is zero.
Chapter 12: Sequences; Induction; the Binomial Theorem
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6. When trying to obtain the cubic and quartic polynomials of best fit, the cubic and quartic terms have coefficient zero and the polynomial of best fit is given as the quadratic in part e. For the exponential function of best fit,
(0.59)(1.83)xy = .
6 7 822.2 40.6 74.2y y y= = =
The sum of these errors becomes quite large. This error shows that the function does not fit the data very well as x gets larger.
7. The quadratic function is best.
8. The data does not appear to be either logarithmic or sinusoidal in shape, so it does not make sense to try to fit one of those functions to the data.