Chapter 12 Sequences; Induction; the Binomial Theorem

Chapter 12: Sequences; Induction; the Binomial Theorem
22.
23.
24.
25.
26.
11 − ⎛ 1 ⎞
d1
= ( − 1) ⎜ ⎟=
1,
⎝21 ⋅ −1⎠
2−1⎛
2 ⎞ 2
d2
= ( − 1) ⎜ ⎟= − ,
⎝2⋅2−1⎠
3
3−1⎛
3 ⎞ 3
d3
= ( − 1) ⎜ ⎟=
,
⎝23 ⋅ −1⎠
5
d
d
4
5
4−1⎛
4 ⎞ 4
= ( − 1) ⎜ ⎟= − ,
⎝2⋅4−1⎠
7
5−1⎛
5 ⎞ 5
= ( − 1) ⎜ ⎟=
⎝25 ⋅ −1⎠
9
1 2
2 2 1 2 4 2
s1 = = = , s
1 2 = = = ,
2
3 + 1 4 2 3 + 1 10 5
s
3 4
2 8 2 2 16 8
= = = , s = = = ,
3 + 1 28 7 3 + 1 82 41
3 3 4 4
s
5 5
5
2 32 8
= = =
3 + 1 244 61
1 2
⎛4⎞ 4 ⎛4⎞
16
s1 = ⎜ ⎟ = , s2
= ⎜ ⎟ = ,
⎝3⎠ 3 ⎝3⎠
9
s
s
3 4
3 4
5
⎛4 ⎞ 64 ⎛4 ⎞ 256
= ⎜ ⎟ = , s = ⎜ ⎟ = ,
⎝3⎠ 27 ⎝3⎠
81
5
⎛4⎞
1024
= ⎜ ⎟ =
⎝3⎠
243
1
( −1) −1 1
t1
= = = − ,
(1+ 1)(1 + 2) 2 ⋅3 6
t
t
t
t
2
3
4
5
2
( −1) 1 1
= = = ,
(2 + 1)(2 + 2) 3⋅4 12
3
( −1) −1 1
= = = − ,
(3 + 1)(3 + 2) 4⋅5 20
4
( −1) 1 1
= = = ,
(4 + 1)(4 + 2) 5⋅6 30
5
( −1) −1 1
= = =−
(5 + 1)(5 + 2) 6⋅7 42
1 2 3
3 3 3 9 3 27
a1 = = = 3, a2 = = , a3
= = = 9,
1 1 2 2 3 3
a
4 5
3 81 3 243
= = , a = =
4 4 5 5
4 5
1 1 2 3 4 5
e e e e e e
27. b1 = = , b
1 2 = , b
2 3 = , b
3 4 = , b
4 5 = 5
28.
2 2 2
1 1 2 2 3 3
1 1 2 3 9
c = = , c = = 1, c = = ,
2 2 2 2 8
2 2
4 16 5 25
c4 = = = 1, c
4 5 = =
5
2 16 2 32
29. Answers may vary. One possibility follows:
Each term is a fraction with the numerator equal
to the term number and the denominator equal to
one more than the term number.
n
an
=
n + 1
30. Answers may vary. One possibility follows:
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
a n =
n n+
1
( )
31. Answers may vary. One possibility follows:
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
a n =
n−1
2
32. Answers may vary. One possibility follows:
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 2 3 .
a n
⎛2
⎞
= ⎜ ⎟
⎝3
⎠
n
33. Answers may vary. One possibility follows:
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 n − 1 as the power.
1
1 n −
= −
a n
( )
1236
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