A simple proof by induction $P(n) = \frac{6^{2n} - 3^n}{11} \in \mathbb{N}$ Verify by induction that $P(n) = \frac{6^{2n} - 3^n}{11} \in \mathbb{N} \quad \forall n \ge 1 \in \mathbb{N}$
Basis: $P(1) \Rightarrow \frac{33}{11} \in \mathbb{N}$.
Induction: if the statement holds for some $n$ $\Rightarrow$ holds for $n+1$
I would appreciate some small hints because I'm stuck at $P(n+1) = \frac{6^{2(n+1)} - 3^{n+1}}{11}$

So the statement $P(n)$ could also be formulated that $11$ divides $6^{2n} -3^n$. So $P(n+1)$ would consider
 \begin{align} 
6^{2(n+1)} - 3^{n+1} & = \\
6^{2n} \times 36 - 3 \times 3^n & = \\
3(12 \times 6^{2n} - 3^n )&= \\
3(11\times 6^{2n} + 6^{2n} - 3^{n}) & = \ldots
\end{align} Can you use $P(n)$ in order to prove $P(n+1)$?
