32n + 2 – 8n – 9 is divisible by 8 for all n ϵ N

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

Let us check for n = 1,

P (1): 3^{2.1 + 2} – 8.1 – 9

: 81 – 17

: 64

P (n) is true for n = 1. Where, P (n) is divisible by 8

Now, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.

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

: 3^{2k + 2} – 8k – 9 = 8λ … (i)

We have to prove,

3^{2k + 4} – 8(k + 1) – 9 is divisible by 8

3^{(2k + 2) + 2} – 8(k + 1) – 9 = 8μ

So,

= 3^{2(k + 1)}.3² – 8(k + 1) – 9

= (8λ + 8k + 9)9 – 8k – 8 – 9

= 72λ + 72k + 81 – 8k – 17 using equation (1)

= 72λ + 64k + 64

= 8(9λ + 8k + 8)

= 8μ

P (n) is true for n = k + 1

Hence, P (n) is true for all n ∈ N.