Chapter 11.7, Problem 2CYP

To determine

To prove that $7^n-1$ is divisible by 6 for all natural numbers n.

Explanation of Solution

Given:

  $7^n-1$

Concept Used:

An identity can be proved using mathematical induction in two steps:

1. First prove the identity is true for $n=1$ .
2. Prove that if it is true for $n=k$ then it is also true for $n=k+1$ .

Calculation:

In order to prove that $7^n-1$ is divisible by 6 for all natural numbers n, use mathematical induction.

First, let’s check if this is true for $n=1$ .

For $n=1$

  $\begin{array}{c}{7}^{\left(1\right)}-1=7-1\\ =6\end{array}$

And 6 is divisible by 6. Thus, the given identity is true for $n=1$ .

Now, let it is true for $n=k;\text{ for some }k>1$ .

Thus, $\left(7^k-1\right)$ is divisible by 6, i.e.

  $7^k-1=6m;\text{ for some integer m}$

Now, let’s show that it is also true for $n=k+1$ , i.e. prove that $\left(7^{k+1}-1\right)$ is also divisible by 6.

Since,

  $\begin{array}{c}{7}^{k+1}-1=7^k\cdot 7-1\\ =\left(7^k-1+1\right)\cdot 7-1\\ =\left(6m+1\right)\cdot 7-1\\ =6m\cdot 7+7-1\\ =6m\cdot 7+6\\ =6\left(7m+1\right)\end{array}$

Thus, $\left(7^{k+1}-1\right)$ is a multiple of 6, i.e. divisible by 6, i.e. the identity is also true for $n=k+1$ , when it is true for $n=k$ .

Thus, by mathematical induction it is proved that $7^n-1$ is divisible by 6 for all natural numbers n .