Question

Show that $3^{2n+2}-8n-9$ is divisible by $8$, where $n$ is any natural number.

Answer 1

We have,

$3^{(2n+2)}-8n-9$

If the number is divisible by 8, then

$8\times 1=8$

$8\times 2=16$

$8\times 3=24$

$8\times 4=32$ and so on.

where n is a natural number.

Let $P\left(n\right)=3^{(2n+2)}-8n-9=8m$,

Where $a\in N$, i.e. a is a natural number.

For n = 1

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

$=3^4-17$

$=81-17$

$=64=8\times 8$

Therefore, $P\left(n\right)$ is true for $n=1$.

Suppose, $P\left(k\right)$ is true.

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

where $a\in N$.

Now, we will prove that $P(k+1)$ is true.

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

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

From equation (1), we get

$=9[8a+9+8k]-8k-17$

$=72a+64k+64$

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

$=8b$

where $b=9a+8k+8$ and $b$ is natural number.

Therefore, $P(k+1)$ is true whenever $P\left(k\right)$ is true.

Therefore, $P\left(n\right)$ is true for $n$, where $n$ is a natural number.

Hence, the given equation is divisible by $8$.