Let the given statement be P(n), i.e., P (n) : $3^{2n+2}$ – 8n – 9 is divisible by 8. First we prove that the statement is true for n = 1, P(1) : $3^{2\cdot 1 + 2}$ – 8⋅1 – 9 = $3^4$ – 17 = 81 – 17 = 64, which is divisible by 8. Assume P(k) is true. i.e., $3^{2k + 2}$ – 8k – 9 = 8g, where g ∈ N ....(i) Now we shall prove that P(k + 1) is true, whenever P(k) is true. For this we have to prove that $3^{2(k + 1) + 2}$ – 8(k + 1) – 9 is divisible by 8. We have, $3^{2(k+1)+2}$ - 8 (k + 1) - 9 = $3^{2k+2+2}$ - 8k - 8 - 9 = $3^{2k+2} \cdot 3^2$ - 8k - 17 = (8g + 8k + 9) . $3^2$ - 8k - 17 [From (i)] = 72g + 72k + 81 – 8k – 17 = 72g + 64k + 64 = 8 (9g + 8k + 8) which shows that $3^{2(k + 1) + 2}$ – 8(k + 1) – 9 is divisible by 8. Hence, P(k + 1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction P(n) is true ∀ n ∈ N.