Let P(n) = 11ⁿ⁺² + 12²ⁿ⁺¹

Now, P(n): 11ⁿ⁺² + 12²ⁿ⁺¹ is divisible by 133 for all n ϵ N

Step1:

P(1) = 1331 + 1728 = 3059

Thus, P(1) is divisible by 133

Step2:

Let, P(m) be divisible by 24

Then, 11^m+2 + 12^2m+1 = 133λ, where λ ϵ N.

Now, we need to show that P(m+1) is true whenever P(m) is true.

So, P(m+1) = 11^m+3 + 12^2m+3

= 11^m+2.11+ 12^2m+1.12²+11.12^2m+1 – 11.12^2m+1

= 11.(11^m+2+ 12^2m+1) + 12^2m+1(144-11)

= 11.133λ + 12^2m+1.133

= 133.(11 λ + 12^2m+1)

Thus, P(m+1) is true.

So, by the principle of mathematical induction, P(n) is true for all n ϵN.