Chapter 12.5, Problem 20E

To determine

To Prove: $3^{2n}-1$ is divisible by 8 for all natural numbers $n$ .

Explanation of Solution

Given information:

Use Mathematical induction to prove the above mentioned formula.

Concept used:

1.By using Mathematical induction technique we need to prove that this formula holds True for all natural numbers. The first step is to prove True for $n=1$ .

2.Second step by using induction hypothesis of mathematical induction we assume for $n=k$ given formula is True.

3.Third step is we need to prove that above formula is True for $n=k+1$ by using formula is True for $n=k$ .

Proof:

First, we need to prove the above formula is True for $n=1$ .

Given formula is

  $3^{2n}-1$ is divisible by 8

Put $n=1$

We get

  $\begin{array}{l}{3}^{2n}-1=3^{2\left(1\right)}-1\\ \Rightarrow 3^{2n}-1=3^2-1\\ \Rightarrow 3^{2n}-1=9-1\\ \Rightarrow 3^{2n}-1=8\end{array}$

8 is divisible by 8.

It True for $n=1$ .

Let us assume it is True for $n=k$

So

  $3^{2k}-1$ is divisible by 8 is True

We need to prove it is True for $n=k+1$

  $\begin{array}{l}{3}^{2n}-1=3^{2\left(k+1\right)}-1\\ \Rightarrow 3^{2n}-1=3^{2k+2}-1\\ \Rightarrow 3^{2n}-1=9\left(3^{2k}\right)-1\\ \Rightarrow 3^{2n}-1=9\left(3^{2k}\right)-9+8\\ \Rightarrow 3^{2n}-1=9\left(3^{2k}-1\right)+8\end{array}$

Here from our assumption $3^{2k}-1$ is divisible by 8,

So $9\left(3^{2k}-1\right)$ is also divisible by 8

So

  $3^{2n}-1$ for $n=k+1$ is divisible by 8.

And hence the proof by Mathematical induction.