Let the given statement be P(n), i.e.,
P (n) : 32n+2 – 8n – 9 is divisible by 8.
First we prove that the statement is true for
n = 1, P(1) : 32⋅1+2 – 8⋅1 – 9 = 34 – 17 = 81 – 17 = 64, which is divisible by 8.
Assume P(k) is true.
i.e., 32k+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
32(k+1)+2 – 8(k + 1) – 9 is divisible by 8.
We have, 32(k+1)+2 - 8 (k + 1) - 9 = 32k+2+2 - 8k - 8 - 9 = 32k+2.32 - 8k - 17
= (8g + 8k + 9) . 32 - 8k - 17 [From (i)]
= 72g + 72k + 81 – 8k – 17 = 72g + 64k + 64 = 8 (9g + 8k + 8)
which shows that 32(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.