Proving that $19\mid 5^{2n+1}+3^{n+2} \cdot 2^{n-1}$ How can I prove that $$5^{2n+1}+3^{n+2} \cdot 2^{n-1} $$ can be divided by 19 for any nonnegative n? What modulo should I choose?

For $n=0$, the formula says that $\left.19\middle|\frac{19}2\right.$, which is false. So consider $n\ge1$:
$$
\begin{align}
5^{2n+1}+3^{n+2}2^{n-1}
&=125\cdot25^{n-1}+27\cdot6^{n-1}\\
&\equiv11\cdot6^{n-1}+8\cdot6^{n-1}&\pmod{19}\\
&=19\cdot6^{n-1}\\
&\equiv0&\pmod{19}
\end{align}
$$
Since
$$
\begin{align}
125&\equiv11&\pmod{19}\\
25&\equiv6&\pmod{19}\\
27&\equiv8&\pmod{19}
\end{align}
$$
