Proving a tricky divisibility by 8

Question

I've been trying to prove by induction that $5^n + 2 * 3^{n + 1} + 1$, for $n > 0$.

My first attempt was trying to prove that the difference $(5^n + 2 * 3^{n + 1} + 1) - (5^{n + 1} + 2 * 3^{n + 2} + 1)$ is always a multiple of 8.

Working this difference, I've found that the difference between consecutive terms is $4(5^{n + 1} + 3^n)$, which proves that their difference is always a multiple of 4. I could isolate an 8 instead of 4, but the inner calculations wouldn't always be integer, which I think doesn't prove anything.

Any tips on how to proceed ?

Answer 1

let $$T_n=5^n+2\cdot3^{n+1}+1$$ and $$T_{n+1}=5^{n+1}+2\cdot 3^{n+1}+1$$ and we have $$T_{n+1}-T_n=4(5^n+3^{n+1})$$ and the right-hande side is divisible by $8$, thus $$T_{n+1}$$ is also divisible by $8$

Answer 2

Suppose you assume truth for $\;n\;$ and now you want to show for $\;n+1\;$ :

$$\begin{align*}&5^{n+1}+2\cdot3^{n+2}+1=3\left(5^n+2\cdot3^{n+1}+1\right)+\overbrace{2\cdot5^n-2}^{=2(5^n-1)}=\\{}\\ &=3\left(5^n+2\cdot3^{n+1}+1)+2\cdot(5-1)(5^{n-1}+5^{n-2}+\ldots+5+1\right)\end{align*}$$

and it is now trivial to check both summands above are multiples of $\;8\;$ ...