Let the given statement be P(n), i.e., P (n) : $41^n-14^n$ is a multiple of 27. First we prove that the statement is true for n = 1, P(1) : 41 – 14 = 27, which is a multiple of 27. Assume P(k) is true i.e., $41^k-14^k$ = 27g, where g ∈ N ... (i) Now prove that P(k + 1) is true. For this we have to prove that $41^(k+1)-14^(k+1)$ is a multiple of 27. Let us consider, $41^(k+1)-14^(k+1)$ = $41^k . 41 - 14^{k+1}$ = (27g + $14^k$)·41 – $14^{k + 1}$ (From (i)) = 27·41g + 41·$14^k$ – $14^{k + 1}$ = 27·41g + $14^k$ [41 – 14] = 27·41g + 27·$14^k$ = 27(41g + $14^k$) ∴ $41^{k + 1} – 14^k$ + 1 is a multiple of 27. Hence, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction P(n) is true ∀ n ∈ N.