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.

I assume that you are trying to prove that the difference is divisible by $512$, as the expression is not.

For simplicity, let $f(n) = 8(5)^{n} + 6(3)^{2n} - 80n - 40$. We can simplify this as $$f(n) = 8(5)^{n} + 2(3)^{2n + 1} - 40(2n + 1).$$
We are trying to prove that $512$ divides $f(n) - f(n - 1)$. We see that it is
$$\left[8(5)^{n} + 2(3)^{2n + 1} - 40(2n + 1)\right] - \left[8(5)^{n - 1} + 2(3)^{2n - 1} - 40(2n - 1)\right] \\ 8(5)^{n} - 8(5)^{n - 1} + 2(3)^{2n + 1} - 2(3)^{2n - 1} - 40(2n + 1) + 40(2n - 1) \\ 8(5)^{n - 1}(5 - 1) + 2(3)^{2n - 1}(3^{2} - 1) - 80 \\ 32(5)^{n - 1} + 16(3)^{2n - 1} - 80 \\ 16(2(5)^{n - 1} + (3)^{2n - 1} - 5)$$
Because we factored $16$, we see that $512/16 = 32$. Hence, we need to prove that $32$ divides $2(5)^{n - 1} + (3)^{2n - 1} - 5$. Let this expression be $g(n)$.
The condition $n \in \mathbb{N}$ tells us that the base case should be $n = 1$. We see that $g(1) = 0$ which is trivial.
Can you do the inducting step?
