   Chapter 2.3, Problem 43E

Chapter
Section
Textbook Problem

In Exercises 35 − 48 , use mathematical induction to prove that the given statement is true for all positive integers n . 5 is a factor of 7 n − 2 n .

To determine

To prove: 5 is a factor of 7n2n for all positive integers n.

Explanation

Formula used:

Mathematical Induction:

The given statement Pn is true for all positive integers n if,

a. Pn is true for n=1

b. The truth of Pk always implies that Pk+1 is true.

Proof:

Consider the statement 5 is a factor of 7n2n for all positive integers n.

By mathematical induction,

a. For n=1,

As, 7n2n=7121=5.

Since, 5 is a factor of 5.

Thus, 5 is a factor of 7n2n for n=1.

Therefore, the statement is true for n=1.

b. Assume that the statement is true for n=k.

That is, 5 is a factor of 7k2k.

7k2k=5z, for some integer z.

To show the statement is true for n=k+1.

Now,

7k+12k+1

=(7k7)(2k2)

By subtracting and adding 7k2,

=(7k

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