Question:
If $\mathrm{P}(n): 2 \times 4^{2 n+1}+3^{3 n+1}$ is divisible by $\lambda$ for all $n \in \mathbf{N}$ is true, then find the value of $\lambda$.
Solution:
For $n=1$,
$\mathrm{P}(1)=2 \times 4^{2+1}+3^{3+1}=2 \times 4^{3}+3^{4}=128+81=209$
For $n=2$,
$\mathrm{P}(2)=2 \times 4^{4+1}+3^{6+1}=2 \times 4^{5}+3^{7}=2048+2187=4235$
As, $\mathrm{HCF}(209,4235)=11$
So, $2 \times 4^{2 n+1}+3^{3 n+1}$ is divisible by 11 .
Hence, the value of $\lambda$ is 11 .