Prove the following by using the principle of mathematical induction for all n ∈ N: 32n + 2 – 8n – 9 is divisible by 8.

Solution:

Let the given statement be P(n), i.e.,

$\mathrm{P}(n): 3^{2 n+2}-8 n-9$ is divisible by 8

It can be observed that $P(n)$ is true for $n=1$ since $3^{2} \times 1+2-8 \times 1-9=64$, which is divisible by 8 .

Let $P(k)$ be true for some positive integer $k$, i.e.,

$3^{2 k+2}-8 k-9$ is divisible by 8

 

$\therefore 3^{2 k+2}-8 k-9=8 m ;$ where $m \in \mathbf{N}$ (1)

We shall now prove that P(k + 1) is true whenever P(k) is true.

Consider

$3^{2(k+1)+2}-8(k+1)-9$

$=3^{2 k+2} \cdot 3^{2}-8 k-8-9$

$=3^{2}\left(3^{2 k+2}-8 k-9+8 k+9\right)-8 k-17$

 

$=3^{2}\left(3^{2 k+2}-8 k-9\right)+3^{2}(8 k+9)-8 k-17$

$=9.8 m+9(8 k+9)-8 k-17$

$=9.8 m+72 k+81-8 k-17$

$=9.8 m+64 k+64$

 

$=8(9 m+8 k+8)$

$=8 r$, where $r=(9 m+8 k+8)$ is a natural number

Therefore, $3^{2(k+1)+2}-8(k+1)-9$ is divisible by 8 .

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.