32n + 7 is divisible by 8 for all n ϵ N

Let P (n): 3²ⁿ + 7 is divisible by 8

Let us check for n = 1,

P (1): 3² + 7 = 9 + 7 = 16

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 + 7 is divisible by 8

: 3^2k + 7 = 8λ

: 3^2k = 8λ – 7 … (i)

We have to prove,

3^{2(k + 1)} + 7 is divisible by 8

3^{2k + 2} + 7 = 8μ

So,

= 3^{2(k + 1)} + 7

= 3^2k.3² + 7

= 9.3^2k + 7

= 9.(8λ – 7) + 7 by using equation (i)

= 72λ – 63 + 7

= 72λ – 56

= 8(9λ – 7)

= 8μ

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

Hence, P (n) is true for all n ∈ N.