Question 22:

Prove the following by using the principle of mathematical induction for all n ∈ N: 3^{2n + 2} – 8n – 9 is divisible by 8.

Answer

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

It can be observed that P(n) is true for n = 1 since 3^{2 × 1 + 2} − 8 × 1 – 9 = 64, which is divisible by 8.

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

i.e., 3^{2k + 2} − 8k − 9 is divisible by 8.

∴3^{2k + 2} − 8k − 9 = 8m; where m ∈ N … (1)

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

Consider

3^2(k+1)+2 – 8(k + 1) – 9

=3^2k+2 ∙ 3² – 8k – 8 – 9

=3²(3^2k+2 – 8k – 9 + 8k + 9) – 8k – 17

= 3²(3^2k+2 – 8k – 9) + 3²(8k + 9) – 8k – 17

= 9.8 m + 9(8k + 9) – 8k – 17

= 9.8m + 72k + 81 – 8k – 17

= 9.8m + 72k + 81 – 8k – 17

= 9.8m + 64k + 64

=8(9m + 8k + 8)

=8r, where r =(9m + 8k + 8) is a natural number

Therefore, 3^2k+2 – 8(k + 1) – 9 is divisible by 8.

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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