Let the given statement be P(n), i.e.,
P (n) : 41n−14n 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.,
41k−14k = 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) = 41k.41−14k+1
= (27g + 14k)·41 – 14k+1 (From (i))
= 27·41g + 41·14k – 14k+1 = 27·41g + 14k [41 – 14]
= 27·41g + 27·14k = 27(41g + 14k)
∴ 41k+1–14k + 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.