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,
$P\left(n\right) = 10^{n} + 3\cdot4^{n+2}+ 5$ 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;
$10^{k+1} + 3\cdot4^{k+3} + 5$ is divisible by $9$
$ P\left(k + 1\right): 10^{k+1}+3\cdot4^{k+3}+5 = 10^{k}\cdot10 + 3\cdot4^{k+3}+5$
$ = \left[9\lambda- 3\left(4^{k+2}\right) - 5\right] \times 10 + 3\cdot4^{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 $n \in N$.