Science:Math Exam Resources/Courses/MATH220/December 2011/Question 06 (b)/Solution 1

First, we prove the claim true when $n=1$. In this case,

$\displaystyle 12^{2n-1}+11^{n+1}=12^{2(1)-1}+11^{1+1}=12+121=133$

which is clearly divisible by 133. We assume the claim is true for $n=k$. For $n=k+1$, we have

${\begin{aligned}12^{2(k+1)-1}+11^{(k+1)+1}&=12^{2k+1}+11^{k+2}\\&=12^{2}\cdot 12^{2k-1}+11\cdot 11^{k+1}\\&=144\cdot 12^{2k-1}+11\cdot 11^{k+1}\\&=(133+11)12^{2k-1}+11\cdot 11^{k+1}\\&=133\cdot 12^{2k-1}+11\cdot 12^{2k-1}+11\cdot 11^{k+1}\\&=133\cdot 12^{2k-1}+11(12^{2k-1}+11^{k+1})\end{aligned}}$

Now, in the last line, the first summand is divisible by 133 and the second summand is also divisible by 133 by the induction hypothesis. Hence we must have that $12^{2(k+1)-1}+11^{(k+1)+1}$ is divisible by 133 as required.