Question

Prove the following by using the principle of mathematical induction for all $n\in N:3^{2n+2}-8n-9$ is divisible by $8$.

Answer 1

Let the given statement be $P\left(n\right)$, i.e.,
$P\left(n\right):3^{2n+2}-8n-9$ is divisible by $8$.
$P\left(n\right)$ is true for $n=1$ since $3^{2\times 1+2}-8\times 1-9=64$, which is divisible by $8$.
Let $P\left(k\right)$ be true for some $k\in N$, i.e.,
$3^{2k+2}-8k-9$ is divisible by $8$
$3^{2k+2}-8k-9=8m$; where $m\in N........\left(i\right)$
We shall now prove that $P(k+1)$ is true whenever $P\left(k\right)$ is true.
Consider $3^{2(k+1)+2}-8(k+1)-9$
$=3^{2k+2}.3^2-8k-8-9$
$=3^2(3^{3k+2}-8k-9+8k+9)-8k-17$
$=3^2(3^{2k+2}-8k-9)+3^2(8k+9)-8k-17$
$=9.8m+9(8k+9)-8k-17$
$=9.8m+72k+81-8k-17$
$=9.8m+64k+64$
$=8(9m+8k+8)$
$=8r$, where $r=(9m+8k+8)$ is a natural number
Therefore, $3^{2(k+1)+2}-8(k+1)-9$ is divisible by $8$.
Thus, $P(k+1)$ is true whenever $P\left(k\right)$ is true.
Hence, by the principle of mathematical induction, statement $P\left(n\right)$ is true for all natural numbers i.e., $n$