**Question:**

Using the principle of mathematical induction, prove each of the following for all $\mathbf{n} \in \mathbf{N}:$

$\left(3^{2 n+2}-8 n-9\right)$ is divisible by 8

**Solution:**

To Prove:

$3^{2 n+2}-8 n-9$ is a divisible of 8

Let us prove this question by principle of mathematical induction (PMI) for all natural numbers

Let $\mathrm{P}(\mathrm{n}):^{2 n+2}-8 n-9$ is a divisible of 8

For n = 1 P(n) is true since

$3^{2 n+2}-8 n-9=3^{2+2}-8 \times 1-9=81-17=64$

Which is divisible of 8

Assume P(k) is true for some positive integer k , ie,

$=3^{2 k+2}-8 k-9$ is a divisible of 8

$\therefore 3^{2 k+2}-8 k-9=m \times 8$, where $m \in \mathrm{N} \ldots(1)$

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

Consider,

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

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

$=3^{2}\left(3^{2(k+1)}-8 k-9+8 k+9\right)-8 k-8-9$

[Adding and subtracting 8k + 9]

$=3^{2}\left(3^{2(k+1)}-8 k-9\right)+3^{2}(8 k+9)-8 k-17$

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

$=9(8 m)+72 k+81-8 k-17[$ Using 1$]$

$=9(8 m)+64 k+64$

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

$=8 \times r$, where $r=9 m+8 k+8$ is a natural number

Therefore $3^{2 k+2}-8 k-9$ is a divisible of 8

Therefore, P (k + 1) is true whenever P(k) is true

By the principle of mathematical induction, P(n) is true for all natural numbers, ie, N.

Hence proved.