**Question:**

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

$\left\{(41)^{n}-(14)^{n}\right\}$ is divisible by 27

**Solution:**

To Prove:

$41^{n}-14^{n}$ is a divisible of 27

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

Let $\mathrm{P}(\mathrm{n}): 41^{n}-14^{n}$ is a divisible of 27

For $n=1 P(n)$ is true since $41^{n}-14^{n}=41^{1}-14^{1}=27$

Which is multiple of 27

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

$=41^{n}-14^{n}$ is a divisible of 27

$\therefore 41^{k}-14^{k}=m \times 27$, where $m \in N \ldots(1)$

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

Consider

$=41^{k+1}-14^{k+1}$

$=41^{k} \times 41-14^{k} \times 14$

$=41\left(41^{k}-14^{k}+14^{k}\right)-14^{k} \times 14$ [Adding and subtracting $14^{k}$ ]

$=41\left(41^{k}-14^{k}\right)+41 \times 14^{k}-14^{k} \times 14$

$=41(27 \mathrm{~m})+14^{k}(41-14)[$ Using 1$]$

$=41(27 \mathrm{~m})+14^{k}(27)$

$=27\left(41 m+14^{k}\right)$

$=27 \times r$, where $r=\left(41 m+14^{k}\right)$ is a natural number

Therefore $41^{k+1}-14^{k+1}$ is divisible of 27

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.