9.1 SOLUTIONS
633
CHAPTER NINE
Solutions for Section 9.1
Exercises
1.
The first term is
2
1
+ 1 = 3
. The second term is
2
2
+ 1 = 5
. The third term is
2
3
+ 1 = 9
, the fourth is
2
4
+ 1 = 17
, and
the fifth is
2
5
+ 1 = 33
. The first five terms are
3
,
5
,
9
,
17
,
33
.
2.
The first term is
1 + (

1)
1
= 1

1 = 0
. The second term is
2 + (

1)
2
= 2 + 1 = 3
. The third term is
3

1 = 2
and
the fourth is
4 + 1 = 5
. The first five terms are
0
,
3
,
2
,
5
,
4
.
3.
The first term is
2
·
1
/
(2
·
1 + 1) = 2
/
3
. The second term is
2
·
2
/
(2
·
2 + 1) = 4
/
5
. The first five terms are
2
/
3
,
4
/
5
,
6
/
7
,
8
/
9
,
10
/
11
.
4.
The first term is
(

1)
1
(1
/
2)
1
=

1
/
2
. The second term is
(

1)
2
(1
/
2)
2
= 1
/
4
. The first five terms are

1
/
2
,
1
/
4
,

1
/
8
,
1
/
16
,

1
/
32
.
5.
The first term is
(

1)
2
(1
/
2)
0
= 1
. The second term is
(

1)
3
(1
/
2)
1
=

1
/
2
. The first five terms are
1
,

1
/
2
,
1
/
4
,

1
/
8
,
1
/
16
.
6.
The first term is
(1

1
/
(1 + 1))
(1+1)
= (1
/
2)
2
. The second term is
(1

1
/
3)
3
= (2
/
3)
3
. The first five terms are
(1
/
2)
2
,
(2
/
3)
3
,
(3
/
4)
4
,
(4
/
5)
5
,
(5
/
6)
6
.
7.
The terms look like powers of
2
so we guess
s
n
= 2
n
. This makes the first term
2
1
= 2
rather than
4
. We try instead
s
n
= 2
n
+1
. If we now check, we get the terms
4
,
8
,
16
,
32
,
64
, . . .
, which is right.
8.
We compare with positive powers of
2
, which are
2
,
4
,
8
,
16
,
32
, . . .
. Each term is one less, so we take
s
n
= 2
n

1
.
9.
We observe that if we subtract
1
from each term of the sequence, we get
1
,
4
,
9
,
16
,
25
, . . .
, namely the squares
1
2
,
2
2
,
3
2
,
4
2
,
5
2
, . . .
.
Thus
s
n
=
n
2
+ 1
.
10.
First notice that
s
n
= 2
n

1
is a formula for the general term of the sequence
1
,
3
,
5
,
7
,
9
,
. . . .
To obtain the alternating signs in the original sequence, we try multiplying by
(

1)
n
. However, checking
(

1)
n
(2
n

1)
for
n
= 1
,
2
,
3
, . . .
gives

1
,
3
,

5
,
7
,

9
,
. . . .
To get the correct signs, we multiply by
(

1)
n
+1
and take
s
n
= (

1)
n
+1
(2
n

1)
.
11.
The numerator is
n
. The denominator is then
2
n
+ 1
, so
s
n
=
n/
(2
n
+ 1)
.
12.
The denominators are the even numbers, so we try
s
n
= 1
/
(2
n
)
. To get the signs to alternate, we try multiplying by
(

1)
n
. That gives

1
/
2
,
1
/
4
,

1
/
6
,
1
/
8
,

1
/
10
, . . . ,
so we multiply by
(

1)
n
+1
instead. Thus
s
n
= (

1)
n
+1
/
(2
n
)
.
13.
We have
s
2
=
s
1
+ 2 = 3
and
s
3
=
s
2
+ 3 = 6
. Continuing, we get
1
,
3
,
6
,
10
,
15
,
21
.
14.
We have
s
2
= 2
s
1
+ 3 = 2
·
1 + 3 = 5
and
s
3
= 2
s
2
+ 3 = 2
·
5 + 3 = 13
. Continuing, we get
1
,
5
,
13
,
29
,
61
,
125
.