Question

By principle of mathematical induction $3^{2n+2}-8n-9$ is divisible by

• A. $9$
• B. $8$
• C. $7$
• D. $6$

Answer 1

Answer: (B)

Let $P\left(n\right)$ be the statement given by
$P\left(n\right);3^{2n+2}-8n-9$ is divisible by $8$.
For $n=1$, we get
$P\left(1\right):3^{2\times 1+2}-\left(8\times 1\right)-9=64=8\times 8$
which is divisible by $8$.
Let $P\left(k\right)$ be true i.e., $3^{2k+2}-8k-9=8\lambda ...\left(i\right)$
For $n=k+1$, we have
$3^{2\left(k+1\right)+2}-8\left(k+1\right)-9$
$=3^{2k+2+2}-8k-8-9=3^{2k+2}{3}^2-8k-17$
$=\left(8\lambda +8k+9\right)3^2-8k-17$[using $\left(i\right)$]
$=\left(8\lambda +8k+9\right)9-8k-17$
$=72\lambda +72k+81-8k-17$
$=72\lambda +64k+64=8\left(9\lambda +8k+8\right)$
which is divisible by $8$.
Therefore, $P\left(k+1\right)$ is true when $P\left(k\right)$ is true.
Hence, from the principle of mathematical induction, the statement is true for all natural numbers $n$.