Chapter 12 Sequences; Induction; the Binomial Theorem

Chapter 12: Sequences; Induction; the Binomial Theorem
10.
11.
12.
13.
n+
1
⎛5
⎞
⎜ ⎟
n+−
1 n
2 ⎛5⎞
5
r =
⎝ ⎠
=
n ⎜ ⎟ =
⎛5
⎞ ⎝2⎠
2
⎜ ⎟
⎝2
⎠
The ratio of consecutive terms is constant,
therefore the sequence is geometric.
1 2
⎛5⎞ 5 ⎛5⎞
25
b1 = ⎜ ⎟ = , b2
= ⎜ ⎟ = ,
⎝2⎠ 2 ⎝2⎠
4
3 4
⎛5 ⎞ 125 ⎛5 ⎞ 625
b3 = ⎜ ⎟ = , b4
= ⎜ ⎟ =
⎝2⎠ 8 ⎝2⎠
16
⎛
n+−
11
2 ⎞
⎜
4 ⎟ n
2 n−( n−1)
r =
⎝ ⎠
= = 2 = 2
n−1
n−1
⎛2
⎞ 2
⎜
4 ⎟
⎝ ⎠
The ratio of consecutive terms is constant,
therefore the sequence is geometric.
11 − 0
2 2 −2
1
c1 = = = 2 = ,
2
4 2 4
2−1 1
2 2 −1
1
c2 = = = 2 = ,
2
4 2 2
3−1 2
2 2
3 2
c
c
= = = 1,
4 2
4−1 3
2 2
4 2
= = = 2
4 2
⎛
n+
1
3 ⎞
⎜
9 ⎟ n+
1
3 n+−
1 n
r =
⎝ ⎠
= = 3 = 3
n n
⎛3
⎞ 3
⎜
9 ⎟
⎝ ⎠
The ratio of consecutive terms is constant,
therefore the sequence is geometric.
1 2
1 d2
3 1 3 9
d = = , = = = 1,
9 3 9 9
3 4
3 27 3 81
d3 = = = 3, d4
= = = 9
9 9 9 9
⎛n+
1⎞
⎜ ⎟
3
⎛n+
1 n⎞
⎝ ⎠ ⎜ − ⎟
⎝ 3 3⎠
⎛n
⎞
⎜ ⎟
⎝3
⎠
2
r = = 2 = 2
1/3
2
The ratio of consecutive terms is constant,
therefore the sequence is geometric.
1/3 2/3 3/3 4/3
1 = 2 , 2 = 2 , 3 = 2 = 2, 4 = 2
e e e e
14.
15.
16.
17.
18.
19.
20.
2( n+
1)
3
2n+ 2−2n
2
r = = 3 = 3 = 9
2n
3
The ratio of consecutive terms is constant,
therefore the sequence is geometric.
21 ⋅
22 ⋅ 4
f1 = 3 = 9, f2
= 3 = 3 = 81,
23 ⋅ 6 24 ⋅ 8
f = 3 = 3 = 729, f = 3 = 3 = 6561
3 4
⎛
n+−
11
3 ⎞
⎜ n+
1
2 ⎟ n n
3 2
r =
⎝ ⎠
= ⋅
n 1 n− 1 n+
1
⎛
−
3 ⎞ 3 2
⎜ n
2 ⎟
⎝ ⎠
n−( n−1) n− ( n+ 1) −1
3
= 3 ⋅ 2 = 3⋅ 2 =
2
The ratio of consecutive terms is constant,
therefore the sequence is geometric.
11 − 0 21 − 1
1 t
1 2 2 2
3 3 1 3 3 3
t = = = , = = = ,
2 2 2 2 2 4
3−1 2 4−1 3
3 3 9 3 3 27
t3 = = = , t
3 3 4 = = =
4 4
2 2 8 2 2 16
⎛
n+
1
2 ⎞
⎜ n+− 11 1 1
3 ⎟ n− n+
3 2
r =
⎝ ⎠
= ⋅
n n n
⎛ 2 ⎞ 3 2
⎜ n−1
3 ⎟
⎝ ⎠
n−− 1 n n+− 1 n −1
2
= 3 ⋅ 2 = 3 ⋅ 2=
3
The ratio of consecutive terms is constant,
therefore the sequence is geometric.
1 2
1 11 − 0
u2
21 −
2 2 2 2 4
u = = = = 2, = = ,
3 3 1 3 3
3 4
2 8 8 2 16 16
u3 = = = , u
3−1 2 4 = = =
4−1 3
3 3 9 3 3 27
a
a
5
n
a
a
5
n
a
a
5
n
n
5−1 4
= 23 ⋅ = 23 ⋅ = 281 ⋅ = 162
1
23 n −
= ⋅
5−1 4
=−2⋅ 4 =−2⋅ 4 =−2⋅ 256 =−512
1
24 n −
=− ⋅
5−1 4
= 5( − 1) = 5( − 1) = 5⋅ 1 = 5
1
5(1) n −
= ⋅ −
5−1 4
a5
= 6( − 2) = 6( − 2) = 6⋅ 16 = 96
1
a 6( 2) n −
= ⋅ −
1254
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