If P(n):

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 .