If $2.4^{2 \mathrm{n}+1}+3^{3 \mathrm{n}+1}$ is divisible by $k$ for all $n \in N$, then $k=$
If $2.4^{2 \mathrm{n}+1}+3^{3 \mathrm{n}+1}$ is divisible by $k$ for all $n \in N$, then $k=$
209
11
8
3
Solution
Let $\mathrm{P}(x)=2.4^{2 \mathrm{n}+1}+3^{3 \mathrm{n}+1}=2^{4 \mathrm{n}+3}+3^{3 \mathrm{n}+1}$
$\begin{aligned}
& \therefore P(1)=2^7+3^4=128+81=209 \\
& P(2)=2^{11}+3^7=2048+2187=4235
\end{aligned}$
H.C.F. of 209 and 435 is 11 So, $\mathrm{P}(x)$ is divisible by 11 .