Prove that for all $n \in N$,$41^{n}-14^{n}$ is a multiple of $27$ using the principle of mathematical induction.

(N/A) Let the given statement be $P(n)$,i.e.,
$P(n): 41^{n}-14^{n}$ is a multiple of $27$.
For $n=1$:
$41^{1}-14^{1} = 27$,which is a multiple of $27$.
Thus,$P(1)$ is true.
Assume $P(k)$ is true for some positive integer $k$,i.e.,
$41^{k}-14^{k} = 27m$ for some $m \in N$ ........$(i)$
We shall now prove that $P(k+1)$ is true whenever $P(k)$ is true.
Consider $41^{k+1}-14^{k+1}$:
$= 41 \cdot 41^{k} - 14 \cdot 14^{k}$
$= 41(41^{k} - 14^{k} + 14^{k}) - 14 \cdot 14^{k}$
$= 41(27m) + 41 \cdot 14^{k} - 14 \cdot 14^{k}$
$= 41 \cdot 27m + 14^{k}(41 - 14)$
$= 41 \cdot 27m + 27 \cdot 14^{k}$
$= 27(41m + 14^{k})$
$= 27r$,where $r = (41m + 14^{k})$ 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,$P(n)$ is true for all $n \in N$.