**Question:**

Using the principle of mathematical induction, prove each of the following for all $\mathbf{n} \in \mathbf{N}:$

$\left(2^{3 n}-1\right)$ is a multiple of 7

**Solution:**

To Prove

$2^{3 n}-1$, which is multiple of 7

Let us prove this question by principle of mathematical induction (PMI) for all natural numbers

$2^{3 n}-1$ is multiple of 7

Let $P(n): 2^{3 n}-1$, which is multiple of 7

For $n=1 P(n)$ is true since $2^{3}-1=8-1=7$, which is multiple of 7

Assume P(k) is true for some positive integer k , ie,

$=2^{3 k}-1=7 m$, where $m \in N \ldots(1)$

We will now prove that P(k + 1) is true whenever P( k ) is true

Consider,

$=2^{3(k+1)}-1$

$=2^{3 k} \times 2^{3}-1$

$=2^{3 k} \times 2^{3}+2^{3}-2^{3}-1$ [Adding and subtracting $\left.2^{3}\right]$

$=2^{3}\left(2^{3 k}-1\right)+2^{3}-1$

$=2^{3}(7 m)+2^{3}-1[$ Using 1]

$=2^{3}(7 m)+7$

$=7\left(2^{3} m+1\right)$

$=7 \times r$, where $r=2^{3} m+1$ is a natural number

Therefore $2^{3 n}-1$ is multiple of 7

Therefore, P (k + 1) is true whenever P(k) is true

By the principle of mathematical induction, P(n) is true for all natural numbers ie, N

Hence proved