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Solved Problems on Introduction to Sequences

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Page 1: Solved Problems on Introduction to Sequences. Solved Problems on Introduction to Sequences by Mika Seppälä Sequences Definition A sequence ( a n )=( a

Solved Problems on Introduction to

Sequences

Page 2: Solved Problems on Introduction to Sequences. Solved Problems on Introduction to Sequences by Mika Seppälä Sequences Definition A sequence ( a n )=( a

Solved Problems on Introduction to Sequences by Mika Seppälä

Sequences

DefinitionDefinition

A sequence (an)=(a1, a2, a3, …) is a rule that assigns number an to every positive integer n.

Page 3: Solved Problems on Introduction to Sequences. Solved Problems on Introduction to Sequences by Mika Seppälä Sequences Definition A sequence ( a n )=( a

Solved Problems on Introduction to Sequences by Mika Seppälä

1

SEQUENCES

1,

1

2,1

6,

1

24,K

⎝⎜⎞

⎠⎟

(3., 3.1, 3.14, 3.141, 3.1415,…)2

3 2,

5

4,10

9,17

16,K

⎝⎜⎞

⎠⎟

Find the general rule defining the following sequences.

Page 4: Solved Problems on Introduction to Sequences. Solved Problems on Introduction to Sequences by Mika Seppälä Sequences Definition A sequence ( a n )=( a

Solved Problems on Introduction to Sequences by Mika Seppälä

4

SEQUENCES

s1=n

Let n be a positive integer. Define the sequence (sk) by

sk+1=3s

k+1, if s

k is odd.

Let n = 3, compute (sk).

Page 5: Solved Problems on Introduction to Sequences. Solved Problems on Introduction to Sequences by Mika Seppälä Sequences Definition A sequence ( a n )=( a

Solved Problems on Introduction to Sequences by Mika Seppälä

1

SEQUENCES

Solution The denominators are

Answer

The general nth term is .

1,

1

2,1

6,

1

24,K

⎝⎜⎞

⎠⎟ = a

n( )

1, 2, 6, 24,…= 1, 2∙1, 3∙2∙1, 4∙3∙2∙1,…

a

n=

1n!

Page 6: Solved Problems on Introduction to Sequences. Solved Problems on Introduction to Sequences by Mika Seppälä Sequences Definition A sequence ( a n )=( a

Solved Problems on Introduction to Sequences by Mika Seppälä

2

SEQUENCES

Solution Remember

π =3.1415926535897932846246….

Answer

(3., 3.1, 3.14, 3.141, 3.1415,…)

The general nth term is n first numbers in the decimal point expansion of π

Page 7: Solved Problems on Introduction to Sequences. Solved Problems on Introduction to Sequences by Mika Seppälä Sequences Definition A sequence ( a n )=( a

Solved Problems on Introduction to Sequences by Mika Seppälä

3

SEQUENCES

Solution Clearly

Answer

= a

n( )

for some f.

a

n=

n2 +1n2

=1 +1n2

.

2,

5

4,10

9,17

16,K

⎝⎜⎞

⎠⎟

an=

f n( ) +1

f n( )f(1) = 1,

f(2) = 22, f(3) = 9 = 32, f(4) = 16 = 42.

Page 8: Solved Problems on Introduction to Sequences. Solved Problems on Introduction to Sequences by Mika Seppälä Sequences Definition A sequence ( a n )=( a

Solved Problems on Introduction to Sequences by Mika Seppälä

4

SEQUENCES

s1=n

Let n be a positive integer. Define the sequence (sk) by

sk+1=3s

k+1, if s

k is odd.

Let n = 3, compute (sk).

Page 9: Solved Problems on Introduction to Sequences. Solved Problems on Introduction to Sequences by Mika Seppälä Sequences Definition A sequence ( a n )=( a

Solved Problems on Introduction to Sequences by Mika Seppälä

4

SEQUENCES

s1=3

sk+1=3s

k+1, if s

k is odd.

Solution s1=3 s2

=3 ⋅s1+1 =10

s

3=

s2

2 =5

s4=3 ⋅s

3+1 =16 s5

=8 s6=4 s7

=2 s8=1

Answer