Chapter 11.7, Problem 5CYU

To determine

To prove that $4^n-1$ is divisible by 3 for all natural numbers n.

Explanation of Solution

Given:

  $4^n-1$

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 $4^n-1$ is divisible by 3 for all natural numbers n, use mathematical induction.

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

For $n=1$

  $\begin{array}{c}{4}^1-1=4-1\\ =3\end{array}$

And 3 is divisible by 3. Thus, the given identity is true for $n=1$ .

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

Thus, $\left(4^k-1\right)$ is divisible by 3, i.e.

  $4^k-1=3m$ , for some integer m .

Now, let’s show that it is also true for $n=k+1$ , i.e. prove that $\left(4^{k+1}-1\right)$ is also divisible by 3. Since,

  $\begin{array}{c}{4}^{k+1}-1=4^k\cdot 4-1\\ =\left(4^k-1+1\right)\cdot 4-1\\ =\left(3m+1\right)\cdot 4-1\because 4^k-1=3m\\ =3m\cdot 4+4-1\\ =3m\cdot 4+3\\ =3\left(4m+1\right)\end{array}$

Thus, $\left(4^{k+1}-1\right)$ is a multiple of 3, i.e. divisible by 3, 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 $4^n-1$ is divisible by 3 for all natural numbers n .