Question

Let $P\left(n\right)={11}^{n+2}+{12}^{2n+1}$, then the least value of the following which $P\left(n\right)$ is divisible by is

• A. $19$
• B. $7$
• C. $133$
• D. none of these

Answer 1

The correct option is B $7$

Given that $P\left(n\right)={11}^{n+2}+{12}^{2n+1}$
Put $n=1$ to obtain $P\left(1\right)={11}^{1+2}+{12}^{2+1}=3059=7\left(437\right)$
Therefore, $P\left(1\right)$ is divisible by $7$
Assume that for $n=k$, $P\left(k\right)={11}^{k+2}+{12}^{2k+1}$ is divisible by $7$
Now, $P(k+1)={11}^{k+3}+{12}^{2k+3}={11.11}^{k+2}+{144.12}^{2k+1}=11({11}^{k+2}+{12}^{2k+1})+133$
$\Rightarrow P(k+1)=11P\left(k\right)+7\left(79\right)$
SInce, $P\left(k\right)$ is divisible by $7$
Therefore, $P(k+1)$ is divisible by $7$
And from the principle of mathematical induction $P\left(n\right)$ is divisible by $7$ for all $n\in N$
Ans: B