Solve the following
Question:

32n+2 −8n − 9 is divisible by 8 for all n ∈ N.

Solution:

Let P(n) be the given statement.

Now,

$P(n): 3^{2 n+2}-8 n-9$ is divisible by 8 for all $n \in N$.

Step :1

$P(1)=3^{2+2}-8-9=81-17=64$

It is divisible by $8 .$

Thus, $P(1)$ is true.

$\operatorname{Step}(2)$ :

Let $P(m)$ be true.

Then, $3^{2 m+2}-8 m-9$ is divisible by 8 .

Let:

$3^{2 m+2}-8 m-9=8 \lambda$ where $\lambda \in N \quad \ldots(1)$

We need to show that $P(m+1)$ is true whenever $P(m)$ is true.

Now,

$P(m+1)=3^{2 m+4}-8(m+1)-17$

$=(8 \lambda+8 m+9)-8 m-8-17 \quad[$ From $(1)]$

$=8 \lambda-16$

$=8(\lambda-1)$

It is divisible by 8 .

Thus, $P(m+1)$ is true.

By the $p$ rinciple of $m$ athematical $i$ nduction, $P(n)$ is true for all $n \in N$.