Question

Show that for any natural number $n,3^{2n+2}-8n-1$ is divisible by $8$.

Answer 1

If a number is divisible by $8$

$16=8\times 2$

$24=8\times 3$

$64=8\times 8$

Any number divisible by $8=8\times$Natural number

Let $P\left(n\right):3^{2n+2}-8n-9=8d$ where $d\in N$

For $n=1$

L.H.S$=3^{2\times 1+2}-8\times 1-9$

$=81-17=64=8\times 8=$R.H.S

$\therefore P\left(n\right)$ is true for $n=1$

Assume $P\left(k\right)$ is true.

$3^{2k+2}-8k-9=8m$ where $m\in N$ .......$\left(1\right)$

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

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

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

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

$=64k+64+72m$

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

$\therefore P(k+1)$ is true whenever $P\left(k\right)$ is true

$\therefore$ By the principle of mathematical induction, $P\left(n\right)$ is true for $n$ where $n$ is a natural number.