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.
 A: So in this question we have to prove that difference of $nth$ and $(n+1)th$ term is divisible by 512.
$$ 8(5^{n+1}-5^n)+6(3^{2n+2}-3^{2n})-80(n+1-n) $$ when divided by $512$ gives remainder $0$. Can you continue it now OP?
Also the attempt which you showed in the question in that you cannot write $n=n+1$. You can write it like $n=x+1$ instead.
A: 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?
A: 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
$5^n+3 \times 9^n$ $\equiv_4$ $1+3 \equiv_4 0$.
