Mathematics - Basic maths Question with Solution | TestHub

Let $P(n)=2\cdot 4^{2n+1}+3^{3n+1}$
Then $P(1)=2.4^3+3^4=209$, which is divisible by 11 but not divisible by 2,7 or 27 .
Further, let $P(k)=2.4^{2k+1}+3^{3k+1}$ is divisible by 11 , that is,
$2.4^{2k+1}+3^{3k+1}=11q$ for some integer q. Now
$\begin{array}{rl}\mathrm{P}(\mathrm{k}+1)& =2\cdot 4^{2\mathrm{k}+3}+3^{3\mathrm{k}+4}\\ & =2\cdot 4^{2\mathrm{k}+1}\cdot 4^2+3^{3\mathrm{k}+1}\cdot 3^3\\ & =16\cdot 2\cdot 4^{2\mathrm{k}+1}+27\cdot 3^{3\mathrm{k}+1}\\ & =16\cdot 2\cdot 4^{2\mathrm{k}+1}+(16+11)\cdot 3^{3\mathrm{k}+1}\\ & =16\left[2\cdot 4^{2\mathrm{k}+1}+3^{3\mathrm{k}+1}\right]+11\cdot 3^{3\mathrm{k}+1}\\ & =16\cdot 11\mathrm{q}+11\cdot 3^{3\mathrm{k}+1}\\ & =11\left(16\mathrm{q}+3^{3\mathrm{k}+1}\right)=11\mathrm{m}\end{array}$
where $m=16q+3^{3k+1}$ is another integer.
$\therefore \mathrm{P}(\mathrm{k}+1)$ is divisible by 11 .