Question

$3^{2n}+7$ is divisible by 8 for all n $ϵ$ N.

Answer 1

Let P(n) : $3^{2m}+7$ is divisible by 8

For n = 1

$3^2+7=16$

Which is divisible by 8

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

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

$3^{2k}+7$ is divisible by 8

$\Rightarrow 3^{2k}+7=8\lambda$ ..........(1)

We have to show that,

$3^{2(k+1)}+7$ is divisible by 8

$3^{2(k+1)}+7=8\mu$

Now,

$3^{2(k+1)}+7$

$=3^{2k}{.3}^2+7$

$=9{.3}^{2k}+7$

$=9.(8\lambda -7)+7$

$=72\lambda -56$

$=72\lambda -56$

$=8(9\lambda -7)$

$=8\mu$

$\Rightarrow$ P(n) is ture for n = k + 1

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