We can write the given statement as

P (n): 3^{2n + 2} – 8n – 9 is divisible by 8

If n = 1 we get

P (1) = 3^{2 × 1 + 2} – 8 × 1 – 9 = 64, which is divisible by 8

Which is true.

Consider P (k) be true for some positive integer k

3^{2k + 2} – 8k – 9 is divisible by 8

3^{2k + 2} – 8k – 9 = 8m, where m ∈ N** **…… (1)

Now let us prove that P (k + 1) is true.

Here

3 ^{2(k + 1) + 2} – 8 (k + 1) – 9

We can write it as

= 3 ^{2k + 2} . 3^{2} – 8k – 8 – 9

By adding and subtracting 8k and 9 we get

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

On further simplification

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

From equation (1) we get

= 9. 8m + 9 (8k + 9) – 8k – 17

By multiplying the terms

= 9. 8m + 72k + 81 – 8k – 17

So we get

= 9. 8m + 64k + 64

By taking out the common terms

= 8 (9m + 8k + 8)

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

So 3 ^{2(k + 1) + 2} – 8 (k + 1) – 9 is divisible by 8

P (k + 1) is true whenever P (k) is true.

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

Answered by Abhisek | 1 year agoGiven an example of a statement P (n) such that it is true for all n ϵ N.

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