Question

$3^{2n+2}-8n-9$ is divisible by 8 for all n $ϵ$ N.

Answer 1

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

For n = 1

$3^{2+2}-8-9$

$=81-17=64$

It is divisible by 8

$\Rightarrow$ P(n) is true for n = 1

Let P(n) is true for n = k, so

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

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

We have to show that,

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

$3^{2(k+1)}{.3}^2-8(k+1)=8\mu$

Now,

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

$=(8\lambda +8k+9)9-8k-8-9$

$=72\lambda +72k+81-8k-17$

$=72\lambda +64k+64$

$=8(9\lambda +8k+8)$

$=8\mu$

$\Rightarrow$ P(n) is true for n = 1

$\Rightarrow$ P(n) is true for all n $ϵ$ N. by PMI.