41ⁿ – 14ⁿis a multiple of 27.

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Solution

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

P(n):41ⁿ – 14ⁿis a multiple of 27.

It can be observed that P(n) is true for n = 1 since , which is a multiple of 27.

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

41^k – 14^kis a multiple of 27

∴41^k – 14^k = 27m, where m  N … (1)

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

Consider

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.

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