CSE 21: Solutions  Problem Set 7
1.
Proof.
Initial step:
For
n
= 1, the lefthand size is 1
3
= 1 and the righthand side is
(
1+1
2
)
2
= 1. Then the
equality holds.
Inductive Hypothesis:
Suppose it is true for
n
=
t
, i.e.,
1
3
+ 2
3
+
· · ·
+
t
3
=
t
+ 1
2
2
.
We prove that it’s true for
n
=
t
+ 1, i.e.,
1
3
+ 2
3
+
· · ·
+
t
3
+ (
t
+ 1)
3
=
t
+ 2
2
2
.
Inductive step:
1
3
+ 2
3
+
· · ·
+
t
3
+ (
t
+ 1)
3
=
(
1
3
+ 2
3
+
· · ·
+
t
3
)
+ (
t
+ 1)
3
=
t
+ 1
2
2
+ (
t
+ 1)
3
Inductive hypothesis
=
(
t
+ 1)
t
2
2
+ (
t
+ 1)
3
=
(
t
+ 1)
2
4
(
t
2
+ 4(
t
+ 1)
)
=
(
t
+ 1)
2
4
(
t
+ 2)
2
=
(
t
+ 2)(
t
+ 1)
2
2
=
t
+ 2
2
2
2. Initial step:
For
n
= 1, the lefthand size is 1
2

1
2
·
0 = 1, and the righthand side is (

1)
2
= 1. So
the equality holds.
Inductive hypothesis:
Suppose it’s true for
n
=
k
, i.e.,
F
2
k

F
k
+1
F
k

1
= (

1)
k
+1
.
We prove that it’s true for
n
=
k
+ 1, i.e.,
F
2
k
+1

F
k
+2
F
k
= (

1)
k
+2
.
1
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Inductive step:
F
2
k
+1

F
k
+2
F
k
=
F
2
k
+1

(
F
k
+1
+
F
k
)
F
k
(by definition)
=
F
2
k
+1

F
k
+1
F
k

F
2
k
=
F
k
+1
(
F
k
+1

F
k
)

F
2
k
=
F
k
+1
F
k

1

F
2
k
(by definition)
=

(

1)
k
+1
(inductive hypothesis)
= (

1)
k
+2
3.
We show three different solutions.
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 Spring '07
 Graham
 Inductive Reasoning, inductive hypothesis

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