MATH 135, Fall 2010
Solution of Assignment #3
Problem 1
.
Prove that
∑
n
i
=1
(2
i

1)
2
=
n
(2
n

1)(2
n
+1)
3
for all
n
≥
1.
Solution.
When
n
= 1 we have
∑
n
i
=1
(2
i

1)
2
= 1
2
= 1 and
n
(2
n

1)(2
n
+1)
3
=
1
·
1
·
3
3
= 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
∑
k
i
=1
(2
i

1)
2
=
k
(2
k

1)(2
k
+1)
3
. Then when
n
=
k
+ 1 we have
=
k
+1
X
i
=1
(2
i

1)
2
=
k
X
i
=1
(2
i

1)
2
+ (2
k
+ 1)
2
=
k
(2
k

1)(2
k
+ 1)
3
+ (2
k
+ 1)
2
= (2
k
+ 1)
k
(2
k

1)
3
+ (2
k
+ 1)
= (2
k
+ 1)
k
(2
k

1) + 3(2
k
+ 1)
3
=
(2
k
+ 1)(2
k
2

k
+ 6
k
+ 3)
3
=
(2
k
+ 1)(2
k
2
+ 5
k
+ 3)
3
=
(2
k
+ 1)(
k
+ 1)(2
k
+ 3)
3
Thus the claim is true when
n
=
k
+ 1. By Mathematical Induction, the claim is true for all
n
≥
0.
Problem 2
.
Let
a
0
= 1 and
a
1
= 1 and for
n
≥
2 let
a
n
= 2
a
n

1
+ 2
a
n

2
. Show that we have
a
n
=
1
2
(
(1 +
√
3)
n
+ (1

√
3)
n
)
for all
n
≥
0.
Solution.
When
n
= 0 we have
a
n
=
a
0
= 1 and
1
2
(
(1 +
√
3)
n
+ (1

√
3)
n
)
=
1
2
(
(1 +
√
3)
0
+ (1

√
3)
0
)
=
1
2
(
1

1
)
= 0, so the claim is true when
n
= 0. When
n
= 1 we have
a
n
=
a
1
= 1 and
1
2
(
(1+
√
3)
n
+(1

√
3)
n
)
=
1
2
(
(1+
√
2)
1
+(1

√
3)
1
)
=
1
2
(
1+
√
3+1

√
3
)
= 1, so the claim is true
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 Spring '10
 Mesta
 Recursion, Inductive Reasoning, Natural number, Mathematical logic, Pallavolo Modena

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