Question

$10^n+3\left(4^{n+2}\right)+5\left(n\in N\right)$ is divisible by

• A. $7$
• B. $5$
• C. $9$
• D. $17$

Answer 1

Answer: (C)

Let $P\left(n\right)$ be the statement given by,
$P\left(n\right)=10^n+3\cdot 4^{n+2}+5$ is divisible by $9$
For $n=1$, we have
$P\left(1\right):10^1+3\cdot \left(4^{1+2}\right)+5=207$, which is divisible by $9$
$\therefore P\left(1\right)$ is true.
Let $P\left(k\right)$ be true. Then, $10^k+3\cdot \left(4^{k+2}\right)+5$ is divisible by $9$
$\Rightarrow 10^k+3\cdot \left(4^{k+3}\right)+5=9\lambda$.
Now, we have to show that $P\left(k+1\right)$ is true, i.e;
$10^{k+1}+3\cdot 4^{k+3}+5$ is divisible by $9$
$P\left(k+1\right):10^{k+1}+3\cdot 4^{k+3}+5=10^k\cdot 10+3\cdot 4^{k+3}+5$
$=\left[9\lambda -3\left(4^{k+2}\right)-5\right]\times 10+3\cdot 4^{k+3}+5$
$=90\lambda -\left(30\times 4^{k+2}\right)-50+\left(3\times 4^{k+3}\right)+5$
$=90\lambda -\left(30\times 4^{k+2}\right)+\left(12\times 4^{k+2}\right)-45$
$=90\lambda -\left(18\times 4^{k+2}\right)-45$
$=9\left(10\lambda -2\times 4^{\text{k+2}}-5\right)$, which is divisible by $9$
$\therefore P\left(k+1\right)$ is true whenever $P\left(k\right)$ is true.
So, $P\left(n\right)$ is true for all $n\in N$.