<s>[INST] Prove that $5^n + 2\cdot3^{n-1} + 1$ is multiple of $8$ Prove that $5^n + 2\cdot3^{n-1}+ 1$ is multiple of $8$.
I've tried using induction (it isn't):
For $n=1$: 
$$5^1 + 2\cdot3^{n-1} + 1 = 8$$
If it is true for $n$, then $n+1$?
\begin{align}
5^{n+1} + 2\cdot3^n + 1 
=
&(4+1)^n\cdot(4+1)+ 2\cdot(2+1)^n + 1 
\\
=& (4^n + n4^{n-1} + 1)\cdot(4+1) + 2\cdot(2^n + n2^{n-1} + 1) + 1 
\\
=
&
(4k+1)\cdot(4+1) + 2(2r+1) + 1 
\\
= &16k+4k+4 +1+4r+2+1 
\\
= 
&20k + 4r + 8 = 4(5+r+2)
\end{align}
But i've only proved it is multiple of $4$.
 [/INST] Hint:
$$
5(5^n+2\cdot 3^{n-1}+1)=5^{n+1}+2\cdot 3^n+1+4(3^{n-1}+1)
$$
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