<s>[INST] Prove that $5$ divides $3^{3n+1}+2^{n+1}$ 
Prove that $5$ divides $3^{3n+1}+2^{n+1}$

I tried to prove the result by induction but I couldn't.
The result is true for $n=1$.
Suppose that the result is true for $n$ i.e $3^{3n+1}+2^{n+1}=5k$ for some $k\in \mathbb{N}$. We study the term 
$$3^{3n+4}+2^{n+2}=3^{3n+1}3^3+2^{n+1}2$$
I tried to prove that that the difference is a multiple of $5$.
$$3^{3n+1}3^3+2^{n+1}2-3^{3n+1}+2^{n+1}=2(3^{3n+1}\cdot 13+2^n)$$
Therefore it's enough to prove that $3^{3n+1}\cdot 13+2^n$ is a multiple of $5$. But if I do again this method applied to this "new problem" is get something similar. I think that there exist a different method to do this using induction.
 [/INST] Or we could just expand and rearrange$$
\begin{align}
3^{3n+1}+2^{n+1} 
&= 3\cdot27^n + 2\cdot2^n \\
&= 3\cdot(25+2)^n +  2\cdot2^n  \\
&= 3\left(5k+2^n \right) + 2\cdot 2^n  \tag{Using Binomial Theorem}\\
&= 5\cdot k'+3\cdot2^n+2\cdot2^n  \\
&= 5\cdot k'+5\cdot 2^n \\
\end{align}
$$
Hence proved
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