Let $P(n): 41^n-14^n$ is a multiple of 27
For $n=1,41^n-14^n=41-14=27$
$\therefore P(n)$ is true for $n=1$
Let $P(k)$ be true.
i.e., $P(k): 41^k-14^k$ is a multiple of 27 .
i.e., $P(k)=41^k-14^k=27 m, m \in N$
or $41^k=27 \mathrm{~m}+14^k \quad$... (i)
We have to prove that $P(k+1)$ is true whenever $P(k)$ is true
$\therefore P(k+1): 41^{k+1}-14^{k+1}=41.41^k-14^{k+1}$
$=41\left(27 m+14^k\right)-14^{k+1} \quad[$ from(i) $]$
$=41.27 m+41.14^k-14^{k+1}$
$=27.41 m+14^k(41-14)$
$=27.41 m+14^k .27=27\left[41 m+14^k\right]$
This shows $41^{k+1}-14^{k+1}$ is a multiple of 27 or $P(k+1)$ is true whenever $P(k)$ is true
Hence, $P(n)$ is true for all $n \in N$