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Solutions to Homework Two
CS 191 :
Discrete Structures I
Jan. 30
, Winter 2007
Due Date: Feb. 13, Tuesday, before the class meeting.
Section 1.7, p60
In Exercises 111, using induction, verify that each equation is true for every positive integer n.
Exercise
(6)
1
3
+ 2
3
+ 3
3
+ … +
n
3
=
+
2
)
1
(
2
n
n
Answer
:
The inductive proof is as follows.
Basis Step.
When
n
= 1, the left side of the equation is equal to 1
3
= 1.
The right side of the equation is equal to
+
2
)
1
1
(
1
2
= 1.
Therefore the equation is true for
n
=1.
Inductive Step.
Assume that the equation is true for
n
=
k
, where
k
≥
1. We shall prove that the equation will be
true for
n
=
k
+1 also. Let
n
=
k
+1. Starting from the left side of the equation, we have the
following derivations:
1
3
+ 2
3
+ 3
3
+ … +
k
3
+ (
k
+1)
3
=
[1
3
+ 2
3
+ 3
3
+ … +
k
3
] + (
k
+1)
3
=
+
2
)
1
(
2
k
k
+ (
k
+1)
3
//by induction!
=
4
)
1
(
2
2
+
k
k
+
4
)
1
(
4
3
+
k
=
4
)]
1
(
4
[
)
1
(
2
2
+
+
+
k
k
k
=
4
)
4
4
(
)
1
(
2
2
+
+
+
k
k
k
1
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4
)
2
(
)
1
(
2
2
+
+
k
k
=
+
+
2
)
2
)(
1
(
2
k
k
Therefore, the equation is also true for
n
=
k
+1.
Inductive proof completes.
In Exercise 2124, use induction to prove the statement.
Exercise
(22)
11
n
– 6 is divisible by 5, for all
n
≥
1.
Answer
:
The inductive proof is as follows.
Basis Step.
When
n
= 1, 11
1
– 6 = 5 which is obviously divisible by 5.
Therefore the statement is true for
n
=1.
Inductive Step.
Assume that the statement is true for
n
=
k
, where
k
≥
1. That is,
11
k
– 6 is divisible by 5.
We shall prove that the statement will be true for
n
=
k
+1 also.
Let
n
=
k
+1. We have the following derivations:
11
n
– 6
= 11
k
+1
– 6
= 11
×
11
k
– 6
= (10+1) 11
k
– 6
= 10
×
11
k
+ 11
k
– 6
= 10
×
11
k
+ (11
k
– 6)
= 5
×
2
×
11
k
+ (11
k
– 6)
Because 5
×
2
×
11
k
is divisible by 5 and by induction, (11
k
– 6) is divisible by 5, 11
k
+1
– 6 is
divisible by 5.
Therefore, the statement is true for
n
=
k
+1 also.
Inductive proof completes.
2
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 Winter '06
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