If P(n) :

Solution:

If $P(n)=2 \times 4^{2 n+1}+3^{3 n+1}$ is divisible by $\lambda \forall n \in \mathbf{N}$ is true

for $n=1$,

$P(1): 2 \times 4^{2(1)+1}+3^{3(1)+1}$

$=2 \times 4^{3}+3^{4}$

$=2 \times 64+81$

$=128+81$

$P(1)=209$

for $n=2$,

$P(2): 2^{3} \times 4^{2(2)+1}+3^{3(2)+1}$

$=2^{3} \times 4^{5}+3^{7}$

$=2^{3} \times 256+2187$

$=2048+2187$

$P(2)=4235$

Since common factor of P(1) and P(2) is 11 

Hence 2 × 42n+1 + 33n+1 is divisible by 11

i.e λ = 11