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CMPSC–360 Discrete Mathematics for Computer Science, Fall 2008
Homework 2 Solutions
Note: These solutions are not necessarily model answers. Rather, they are designed to be tutorial in nature,
and sometimes contain a little more explanation than an ideal solution. Also, bear in mind that there may be
more than one correct solution.
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1. Practice with induction
(a) The base case
,
n
= 2
, is the statement that
5
4
is less than
2

1
2
=
6
4
, which is clearly true.
The inductive hypothesis
(for some integer
n
≥
2
) is that
1 +
1
4
+
. . .
+
1
n
2
<
2

1
n
.
In the inductive step
, we need to prove that
1 +
1
4
+
. . .
+
1
(
n
+1)
2
<
2

1
n
+1
. We do this as follows:
1 +
1
4
+
. . .
+
1
(
n
+ 1)
2
=
p
1 +
1
4
+
. . .
+
1
n
2
P
+
1
(
n
+ 1)
2
<
p
2

1
n
P
+
1
(
n
+ 1)
2
(by induction hypothesis)
= 2

p
1
n

1
(
n
+ 1)
2
P
= 2

p
n
2
+ 2
n
+ 1

n
n
(
n
+ 1)
2
P
= 2

n
2
+
n
n
(
n
+ 1)
2

1
n
(
n
+ 1)
2
= 2

1
n
+ 1

1
n
(
n
+ 1)
2
<
2

1
n
+ 1
.
Thus, by induction, the hypothesis is true for all integers
n
≥
2
.
(b) For the base case
,
n
= 1
, we simply observe that
4
1+1
+ 5
2
·
1

1
= 16 + 5 = 21
, which is divisible by
21
.
Then we assume the inductive hypothesis
, that
4
n
+1
+ 5
2
n

1
is divisible for
21
(for some specific
n
≥
1
).
In the inductive step
we need to prove that
4
n
+2
+5
2
n
+1
is divisible by 21. In order to use the inductive
hypothesis, we want to somehow manipulate this expression so that the expression for
n
appears. We
have
4
n
+2
+ 5
2
n
+1
= 4
·
4
n
+1
+ 25
·
5
2
n

1
= 4
·
4
n
+1
+ (4 + 21)
·
5
2
n

1
= 4(4
n
+1
+ 5
2
n

1
) + 21
·
5
2
n

1
.
[36 points, each worth 12 pts]
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View Full DocumentLooking at the last line, we see that the expression in parentheses is divisible by
21
by the inductive
hypothesis, and obviously the second term is divisible by
21
, so the entire quantity is divisible by
21
,
as desired. Therefore, the proof is complete by induction.
(c) For the base cases
we verify that for
n
= 0
(even) we have
∑
n
i
=0
f
i
2
F
= 0 =
0
2
4
, and also for
n
= 1
we have
∑
n
i
=0
f
i
2
F
= 0 =
1

1
4
. (We use two base cases because we will handle odd and even
n
separately.)
In the induction hypothesis
we assume that the equality holds for
n

2
. In the induction step
we
want to show that it holds for
n
. (Note that we are using strong induction
here, because we are using
P
(
n

2)
rather than
P
(
n

1)
to prove
P
(
n
)
.) To show this, notice that
n
s
i
=0
f
i
2
F
=
n

2
s
i
=0
f
i
2
F
+
f
n

1
2
F
+
f
n
2
F
Thus if
n
is odd (so
n

1
is even), we obtain, using the induction hypothesis and
f
n

1
2
F
=
f
n
2
F
=
n

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 Fall '08
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