Question

Prove the following by the principle of mathematical induction: 

3^{2n + 2} – 8n – 9 is divisible by 8 for all n ϵ N.

Answer 1

Let P(n): 3^{2n + 2} – 8n – 9 

For n = 1 

= 3^{2.1 + 2} - 8.1 - 9 

= 81 – 17 

= 64 

Since, it is divisible by 8 

Let P(n) is true for n=k, so 

= 3^{2k + 2} – 8k – 9 is divisible by 8 

= 3^{2k + 2} – 8k – 9 = 8λ - - - - - (1) 

We have to show that, 

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

= 3^{(2k + 2) + 2} – 8(k + 1) – 9 = 8μ 

Now, 

= 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 

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

Hence, P(n) is true for all n∈N by PMI