Question

If $n\in N$, then ${11}^{n+2}+{12}^{2n+1}$ is divisible by:
(use principle of mathematical induction)

• A. $113$
• B. $123$
• C. $133$
• D. $143$

Answer 1

The correct option is C $133$

Let $S\left(n\right):{11}^{n+2}+{12}^{2n+1}$ is divisible by $p$ is true for $n\in N$
$\Rightarrow S\left(n\right):{11}^{n+2}+{12}^{2n+1}=\lambda p,\lambda \in Z^{+}$
Let $S\left(n\right)$ is true for $k=n$
Now for $k=n+1,$
$S(n+1):{11}^{n+3}+{12}^{2n+3}=11\cdot {11}^{n+2}+144\cdot {12}^{2n+1}=11\cdot {11}^{n+2}+144(\lambda p-{11}^{n+2})=(-133){11}^{n+2}+144\lambda p$
Clearly, $p$ must be multiple of $133,$ for $S(n+1)$ to be true.
So, ${11}^{n+2}+{12}^{2n+1}$ is divisible by $133.$