Question 23:

Prove the following by using the principle of mathematical induction for all n ∈N : 4ⁿ – 14ⁿ is a multiple of 27.

Answer

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 41¹ – 14¹ = 27, which is multiple of 27.

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

41^k – 14^k is 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

41^k+1 – 14^k+1

=41^k ∙ 41 – 14^k ∙ 14

=41(41^k – 14^k + 14^k) – 14^k ∙ 14

=41(41^k – 14^k) + 41∙14^k – 14^k ∙ 14

=41.27m + 14^k(41 – 14)

=41.27m + 27.14^k

=27(41m – 14^k)

= 27 × r, 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, statement P(n) is true for all natural numbers i.e., n.

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