MATH 55
Prof. Bernoff
Fall 2007
Harvey Mudd College
Homework 4 – Solutions
D1:
Scheinerman, Section 21 (p. 168) # 3(b),(d): Prove the following by induction.
(b) We wish to show for
n
a positive integer that
1
3
+ 2
3
+
· · ·
+
n
3
=
n
2
(
n
+ 1)
2
4
.
Solution:
Base case(n=1)
: Note
1
2
(1 + 1)
2
4
=
4
4
= 1 = 1
3
.
Induction Hypothesis
: Suppose
1
3
+ 2
3
+
· · ·
+
k
3
=
k
2
(
k
+ 1)
2
4
,
we wish to show
1
3
+ 2
3
+
· · ·
+ (
k
+ 1)
3
=
(
k
+ 1)
2
(
k
+ 2)
2
4
.
Note that
1
3
+ 2
3
+
· · ·
+
k
3
+ (
k
+ 1)
3
=
k
2
(
k
+ 1)
2
4
+ (
k
+ 1)
3
(IHOP)
=
(
k
+ 1)
2
(
k
2
4
+
k
+ 1)
,
=
(
k
+ 1)
2
(
k
2
+ 4
k
+ 4)
4
,
=
(
k
+ 1)
2
(
k
+ 2)
2
4
,
so the formula follows by the Principle of Mathematical Induction.
(d) We wish to show for
n
a positive integer that
1
1
·
2
+
1
2
·
3
+
· · ·
+
1
n
·
(
n
+ 1)
= 1

1
n
+ 1
.
Solution:
Base case(n=1)
: Note
1

1
1 + 1
=
1
2
=
1
1
·
2
.
Induction Hypothesis
: Supposing
1
1
·
2
+
1
2
·
3
+
· · ·
+
1
k
·
(
k
+ 1)
= 1

1
k
+ 1
,
we wish to show
1
1
·
2
+
1
2
·
3
+
· · ·
+
1
(
k
+ 1)
·
(
k
+ 2)
= 1

1
k
+ 2
.
Note that
1
1
·
2
+
1
2
·
3
+
· · ·
+
1
(
k
+ 1)
·
(
k
+ 2)
=
1

1
k
+ 1
+
1
(
k
+ 1)
·
(
k
+ 2)
(IHOP)
=
1

k
+ 2
(
k
+ 1)
·
(
k
+ 2)
+
1
(
k
+ 1)
·
(
k
+ 2)
,
=
1

k
+ 1
(
k
+ 1)
·
(
k
+ 2)
,
=
1

1
k
+ 2
,
so the formula follows by the Principle of Mathematical Induction.
1
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D2:
Scheinerman, Section 21 (p. 168) # 5. A group of people stand in line to purchase movie tickets. The
first person in line is a woman and the last person in line is a man. Use proof by induction to show that
somewhere in the line a woman is directly in front of a man.
Solution:
We prove the claim inductively:
Base case:
If the line is just those two people, then the man in back is directly behind the woman in front.
Induction Hypothesis:
Supposing now that the statement is true for lines up to
n
persons long. In a line
of length
n
+ 1, if the person directly behind the woman in front is a man, then we are done. Otherwise, the
n
persons behind her constitute a line in which the person in front of a woman and the person in back is a
man. By the inductive hypothesis, somewhere in that subline we must have a woman standing in front of
a man.
Therefore the claim follows from the Principle of Mathematical Induction.
D3:
Scheinerman, Section 21 (p. 168) # 10. Prove that every positive integer can be expressed as the sum
of distinct Fibonacci numbers. For example, 20 = 2 + 5 + 13 where 2
,
5 and 13 are, of course, Fibonacci
numbers. Although we can write 20 = 2 + 5 + 5 + 8, this does not illustrate the result because we have used
5 twice.
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 Fall '07
 Bernoff
 Math, Mathematical Induction, Recursion, Natural number, Fibonacci number, distinct Fibonacci numbers, Initiallly Alice

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