CSCE 235
Homework
Assignment 6 – Solution
Assigned: February 25, 2009
Due: 12:30 p.m. March 4,
2009
(Homework 5 minutes late will not be
accepted)
1. (10
points) Prove
by induction that the following is true for 
:
First, we show that
the basis is true. When
,
.
We have shown that the basis step is true by inspection.
Second,
we show that the induction step is true.
Assume that 
is true (when 
):
Now,
we need to show that 
is true (
) when given that the above is true:
From , we know that
So,
we can rewrite the as:
.
Thus we have shown
that is true. Therefore, by the Principle of Mathematical
Induction, we have shown that
.
2. (10 points) Prove by induction that
for
all 
.
First,
we want to establish the basis. For 
,
So, the basis is true by inspection.
Next, we want to establish the induction step. That is, given the kth proposition is true, (k+1)th proposition is true. Thus, P(k) is:
And we want to show that P(k+1) is true:
To
prove, since we know :
= 
= 
= 
=
=
=
.
Thus, we have shown that the induction step holds: P(k+1) is true when P(k) is true.
Therefore,
by the Principle of Mathematical Induction, we have shown that for all
.
3. (10 points) Prove by induction that 
for all .
First, we want
to show that the basis is true. When ,
.
Thus, we have
shown that the basis is true by inspection.
Second, we want to show that the induction step is
true. That is, given P(k), P(k+1) is true. P(k) is when n = k:
P(k+1) is when n = k+1:
.
To show P(k+1) is true:
Since we know that , we can substitute
this into the above:
.
Now, it is a simple algebraic exercise:
.
Thus, we have
shown that the induction step holds true.
Therefore, by the Principle of Mathematical Induction, we
have shown that for all
.
4. (10
points) Prove
by induction that 
for all 
.
First, we show that
the basis is true. When
,
.
We have shown that the basis step is true by inspection.
Second,
we show that the induction step is true.
Assume that is true (when ):
Now,
we need to show that 
is true (when ) given that the above is true:
.
Given , we have:
.
Thus we have shown
that 
is true. Therefore, by the Principle of Mathematical
Induction, we have shown that
for
all 
.
5. (10
points) Prove
by induction that 
for all 
.
First, we show that
the basis is true. When
,
.
We have shown that the basis step is true by inspection.
Second,
we show that the induction step is true.
Assume that 
is true (when 
):
Now,
we need to show that 
is true (when 
) given that the above is true:
.
Given , we have:
.
Thus we have shown
that is true. Therefore, by the Principle of Mathematical
Induction, we have shown that
for
all .
6.
(10 points) Prove
by induction that 
whenever n is a
positive integer.
First, we show that
the basis is true. When
,
.
We have shown that the basis step is true by inspection.
Second,
we show that the induction step is true.
Assume that is true:
Now,
we need to show that is true given that the
above is true:
.
We
know that 
. From
, we know that
.
Thus we have shown
that is true. Therefore, by the Principle of Mathematical
Induction, we have shown that
whenever
n is a positive integer.
7. (10 points) Prove by induction that all numbers of
the form 
are divisible by 5 for
all .
First, we show that the basis is
true. When , = 
which is divisible by 5. So, the basis is true by inspection.
Now, we want to show that the induction step is true. Our nth proposition is that:
is
divisible by 5.
We want to show that if the above is true, then the (n+1)th proposition is also true:
is
divisible by 7.
Thus, we have 
Since we know that is divisible by 5,
thus, we know that the sum 
is divisible by 5 as
well. And thus, we have shown that is divisible by 5.
Therefore, by the Principle of
Mathematical Induction, we have shown that all numbers of the form are divisible by 5 all .
8. (10 points) The product of n elements 
is denoted 
. Let x be a
real number, 
. Prove by induction
that
for
any integer .
First,
we show that the basis is true. When ,
.
So, the basis is true by inspection.
Now, we want to show that the induction step holds. The nth proposition says that
.
We want to show that given the nth proposition, the (n+1)th proposition is true:
.
So, we have
.
Thus, we have shown that the induction holds.
Therefore, by the Principle of Mathematical Induction, we have shown that
for
any integer .
Problems
based on (Rosen 2003, Gossett 2003, Hein 2002, Goodaire
and Parmenter 2002, Ross and Wright 1988).