MATH 135 Algebra, Solutions to Assignment 3
1:
(a) Let
a
1
= 1 and
a
n
+1
= 3
a
n
+ 2 for
n
≥
1. Show that
a
n
= 2
·
3
n

1

1 for all
n
≥
1.
Solution: We claim that
a
n
= 2
·
3
n

1

1 for all
n
≥
1. When
n
= 1 we have
a
n
=
a
1
= 1 and
2
·
3
n

1

1 = 2
·
3
0

1 = 2
·
1

1 = 1 so the claim is true when
n
= 1. Let
k
≥
1 and suppose the claim is
true when
n
=
k
, that is suppose that
a
k
= 2
·
3
k

1

1. Then when
n
=
k
+ 1 we have
a
n
=
a
k
+1
= 3
a
k
+ 2 = 3
(
2
·
3
k

1

1
)
+ 2 = 2
·
3
k

3 + 2 = 2
·
3
k

1 = 2
·
3
n

1

1
.
Thus the claim is true when
n
=
k
+ 1. By Mathematical Induction,
a
n
= 2
·
3
n

1

1 for all
n
≥
1.
(b) Let
a
1
= 3 and
a
n
+1
= 2
a
n

1 for
n
≥
1. Find a closed form formula for
a
n
.
Solution: Using the given recursion formula, we ﬁnd that
a
1
= 3,
a
2
= 5,
a
3
= 9,
a
4
= 17 and
a
5
= 33.
Notice that
a
n
= 2
n
+ 1 for
n
= 1
,
2
,
3
,
4
,
5. We claim that
a
n
= 2
n
+ 1 for all
n
≥
1. When
n
= 1 the claim
is true. Let
k
≥
1 and suppose that the claim is true when
n
=
k
, that is suppose that
a
k
= 2
k
+ 1. Then
when
n
=
k
+ 1 we have
a
n
=
a
k
+1
= 2
·
a
k

1 = 2(2
k
+ 1)

1 = 2
k
+1
+ 2

1 = 2
k
+1
+ 1 = 2
n
+ 1
.
Thus the claim is true when
n
=
k
+ 1. By Mathematical Induction, we have
a
n
= 2
n
+ 1 for all
n
≥
1.
(c) Let
a
1
= 2 and
a
n
+1
=
5
a
n

4
a
n
for
n
≥
1. Show that 1
≤
a
n
≤
a
n
+1
≤
4 for all
n
≥
1.
Solution: We claim that 1
≤
a
n
≤
a
n
+1
≤
4 for all
n
≥
1. We have
a
1
= 2 and the recursion formula gives
a
2
=
5
a
1

4
a
1
=
5
·
2

4
2
= 3, and so we do have 1
≤
a
1
≤
a
2
≤
4 and so the claim is true when
n
= 1. Let
k
≥
1 and suppose the claim is true when
n
=
k
, that is suppose that 1
≤
a
k
≤
a
k
+1
≤
4. We have
1
≤
a
k
≤
a
k
+1
≤
4 =
⇒
1
≥
1
a
k
≥
1
a
k
+1
≥
1
4
=
⇒
4
≥
4
a
k
≥
4
a
k
+1
≥
1 =
⇒ 
4
≤ 
4
a
k
≤ 
4
a
k
+1
≤ 
1
=
⇒
1
≤
5

4
a
k
≤
5

4
a
k
+1
≤
4 =
⇒
1
≤
5
a
k

4
a
k
≤
5
a
k
+1

4
a
k
+1
≤
4 =
⇒
1
≤
a
k
+1
≤
a
k
+2
≤
4
.
Thus the claim is true when
n
=
k
+ 1. By Mathematical Induction, 1
≤
a
n
≤
a
n
+1
≤
4 for all
n
≥
1.
2:
(a) Let
a
0
= 0 and
a
1
= 1 and for
n
≥
2 let
a
n
=
a
n

1
+6
a
n

2
. Show that
a
n
=
1
5
(
3
n

(

2)
n
)
for all
n
≥
0.
Solution: We claim that
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 Fall '08
 ANDREWCHILDS
 Algebra, Binomial Theorem, Mathematical Induction

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