Question

${11}^{n+2}+{12}^{2n+1}$ is divisible by 133 for all $nϵN$.

Answer 1

Let $P\left(n\right):{11}^{n+2}+{12}^{2n+1}$ is divisible by 133

For n = 1

$={11}^3+{12}^3$

$=1331+1728$

$=3059$

It is divisible of 133

$\Rightarrow$ P(n) is true for n = 1

Let P(n) is true for n = k, so

${11}^{k+2}+{12}^{2k+1}$ is divisible by 133

${11}^{k+2}+{12}^{2k+1}=133\lambda$ ........(1)

We have to show that,

${11}^{k+3}+{12}^{2k+3}$ is divisible by 133

Now,

$={11}^{k+2},11+{12}^{2k+1}.{12}^2$

$=(133\lambda -{12}^{2k+1})11+{12}^{2k+1}.144$

$=11.133\lambda -{11.12}^{2k+1}+{144.12}^{2k+1}$

$=11.133\lambda -{133.12}^{2k+1}$

$=133(11\lambda +{12}^{2k+1})$

$=133\mu$

$\Rightarrow$ P(n) is true for n = k + 1

$\Rightarrow$ P(n) is true for all $nϵN$ by PMI.