Prove that $8(5)^n + 6(3)^{2n} - 80n - 14$ is divisible by $512$ for all $n \in \mathbb{N}$ by induction. Note: question was updated to correct the constant term $-14$ (vs. $-40$).
We need to prove that $8(5)^n + 6(3)^{2n} - 80n - 14$ is divisible by $512$ for all $n \in \mathbb{N}$.
I started by taking $n = 1$, then for $n = k$ and then for $n = k+1$.
I am stuck here. Not able to solve further but still trying. What to do next?
Thanks for your time.

soupless gave a good answer, I do think that, for completeness, it would be instructive to elaborate and show that 32 indeed divides $g(n) \doteq [2 \times 5^{n}]+ [3 \times 9^{n}] - 5$ for $n=1,2,3,\ldots$
This is clearly true for $n=1$. Then to show that $32|g(n)$ for general $n$, it suffices to show that $32|[g(n+1) - g(n)]$. To this end, we use the technique that @soupless used above in his answer, and note the following:
$$g(n+1)-g(n) = (8 \times 5^n) + (24 \times 9^{n})$$ $$ = 8 \times [5^n + (3 \times 9^n)].$$
Then 32 indeed divides $g(n+1)-g(n)$ iff 4 divides $5^n+(3 \times 9^n)$.
But this is easy to see that 4 indeed divides $5^n+3 \times 9^n$; to elaborate, $5^n \equiv_4 1$ and $9^n \equiv_4 1$, so
$5^n+3 \times 9^n$ $\equiv_4$ $1+3 \equiv_4 0$.
