If n is a positive integer, then 2. $4^{2n+1}+3^{3n+1}$ is divisible by : 

Let P(n) = 2. $4^{2n+1}+3^{3n+1}$,

Then P(l) = 2. $4^3+ 3^4$= 209, which is divisible

by 11 but not divisible by 2, 7 or 27. 

Further, let $\mathrm{P}\left(k\right)=2\cdot 4^{2k+1}+3^{3k+1}$ is divisible 

by 11, i.e., ${2.4}^{2k+1}+3^{3k+1}=11q$ for some integer q. 

Now $\mathrm{P}(k+1)=2\cdot 4^{2k+3}+3^{3k+4}$

$\begin{array}{l}=2\cdot 4^{2k+1}\cdot 4^2+3^{3k+1}\cdot 3^3\\ =16\cdot 2\cdot 4^{2k+1}+27\cdot 3^{3k+1}\\ =16\cdot 2\cdot 4^{2k+1}+(16+11)\cdot 3^{3k+1}\\ =16\left[2\cdot 4^{2k+1}+3^{3k+1}\right]+11\cdot 3^{3k+1}\\ =16\cdot 11q+11\cdot 3^{3k+1}\end{array}$

$=11\left(16q+3^{3k+1}\right)=11\mathrm{m}$

where $m=16q+3^{3k+1}$ is another integer. 

$\therefore P(k+1)$ $is divisible by 11.$

$\therefore P\left(n\right)=2\cdot 4^{2n+1}+3^{3n+1}$ is divisible by 11 

for all n $\in$N.