Proving divisibility by $9$ for $10^n + 3 \times 4^{n+2} + 5$. I am trying to prove that for all $n$, $9$ divides $10^n + 3 \times 4^{n+2} + 5$. I first tried induction on $n$, but couldn't get anywhere with the induction step. I then tried to use modular arithmetic. The furthest I could get was:

As $10 \equiv 1 \mod 9$, $10^n \equiv 1^n = 1$ for all $n$, so modulo $9$, we have
\begin{align*}
10^n + 3 \cdot 4^{n+2} + 5 & = 3 \cdot 4^{n+2} + 6 \\
& = 3\left(4^{n+2} + 2\right) \\
& = 3\left(\left(2^2\right)^{n+2} + 2 \right) \\
& = 3\left(2^{2(n+2)} + 2 \right) \\
& = 3\left(2\left(2^{2(n+2) - 1} + 1\right) \right)
\end{align*}

I need to something get a factor of $3$ to show that the entire expression is divisible by $9$ and hence equal to $0$, mod $9$. But, with a sum of even terms, this does not appear possible.
Any hints on how to proceed would be appreciated. Is induction the standard way to prove something like this?

Here is another way if you know the binomial theorem.
Observe that $$10^n = (1+9)^n = 1 + {n \choose1}9 + \ldots + 9^n = 1 + 9A$$ where $A= {n \choose1}+ \ldots + 9^{n-1}$ and similarly $$4^{n+2} = (1+3)^{n+2} = 1 + (n+2) 3 + {{n+2} \choose 2}3^2 + \ldots + 3^{n+2} = 1 + 3(n+2) + 3^2B$$
So $$10^n + 3 \cdot 4^{n+2} + 5 = 1 + 9A + 3( 1 + 3(n+2) + 3^2B ) + 5 = \color{red}{9} {\left(1 + A + (n+2) + 3B \right) }$$
