Prove the following by using the principle of mathematical induction for all n ∈ N: 41n – 14n is a multiple of 27.

Solution:

Let the given statement be P(n), i.e.,

$P(n): 41^{n}-14^{n}$ is a multiple of 27

It can be observed that $P(n)$ is true for $n=1$ since $41^{1}-14^{1}=27$, which is a multiple of 27 .

Let P(k) be true for some positive integer k, i.e.,

$41^{k}-14^{k}$ is a multiple of 27

 

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

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

Consider

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

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

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

$=41\left(41^{k}-14^{k}\right)+41.14^{k}-14^{k} \cdot 14$

$=41.27 m+14^{k}(41-14)$

$=41.27 m+27.14^{k}$

 

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

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

Therefore, $41^{k+1}-14^{k+1}$ is a multiple of 27 .

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.