Question

Explain the method of mathematical induction and use it to show that ${11}^{n+2}+{12}^{2n+1}$ where n is natural number is divisible by 133

Answer 1

Let $A_n={11}^{n+2}{12}^{2n+1}$
The assertion is valid for $n=1$ Since,
$A_1={\left(11\right)}^{1+2}+{\left(12\right)}^{2+1}=3059=133\times 23$
Assume that the assertion holds for n = m ,that is, let
$A_m={11}^{m+2}+{12}^{2m+1}=133K$..............................(1)
where K is a positive integer.
Then $A_{m+1}={11}^{m+3}+{12}^{2(m+1)+1}$
${11}^{m+3}+{12}^{2m+3}$
${11.11}^{m+2}+{144.12}^{2m+1}$
$11(133K-{12}^{2m+1})+{144.12}^{2m+1}$............................by (1)
$11.\left(133K\right)+(144-11){12}^{2m+1}$
$11.133K+133.{12}^{2m+1}$
$133(11K+{12}^{2m+1})$
This shows that $A_{m+1}$ is divisible by 133. Hence by induction, $A_n$ is divisible by 133 for all natural number n.