Answer
Let $P(n)$ denote the proposition: "If $n$ is a positive integer, then $133$ divides $11^{n+1} + 12^{2n-1}$."
$P(1)$ is true, since $11^{(1)+1} +12^{2(1)-1} = 133$.
Assume that $P(k)$ is true for an arbitrary integer $k$. Then
\begin{align*}
11^{(k+1)+1} + 12^{2(k+1)-1} &= 11^{k+2} + 12^{2k+1} \\
&= 11(11^{k+1}) +144(12^{2k-1}) \\
&= 144(11^{k+1} + 12^{2k-1}) - 133(11^{k+1})
\end{align*}
Each of the bracketed terms is divisible by $133$, so $P(k+1)$ is also true. Therefore by method of mathematical induction $P(n)$ is true - $133$ divides $11^{n+1} + 12^{2n-1}$ whenever $n$ is a positive integer.
Work Step by Step
First, we must give the proposition. We let $P(n)$ denote the proposition: "If $n$ is a positive integer, then $133$ divides $11^{n+1} + 12^{2n-1}$."
Next, we must prove our base case $P(1)$ is true by showing that the proposition is true when $n=1$. When $n=1$, $11^{n+1} + 12^{2n-1} = 11^{(1)+1} +12^{2(1)-1} = 121 + 12 =133$. $133$ is divisible by $133$, so therefore $P(1)$ is true.
Finally, we must carry out the inductive step, Assume that $P(k)$ is true; that is, we assume that $133$ divides $11^{k+1} + 12^{2k-1}$ for an arbitrary positive integer $k$. Now we must show that this implies that $P(k+1)$ is also true.
\begin{align*}
11^{(k+1)+1} + 12^{2(k+1)-1} &= 11^{k+2} + 12^{2k+1} \\
&= 11(11^{k+1}) +144(12^{2k-1}) \\
&= 144(11^{k+1}) - 133(11^{k+1}) + 144(12^{2k-1}) \\
&= 144(11^{k+1} + 12^{2k-1}) - 133(11^{k+1})
\end{align*}
The first bracketed term is divisible by $133$, via our inductive hypothesis; the second bracketed term is trivially divisible by $133$. This means that the entire expression must also be divisible by $133$, and that $P(k+1)$ is true. Therefore by method of mathematical induction $P(n)$ is true - $133$ divides $11^{n+1} + 12^{2n-1}$ whenever $n$ is a positive integer.