Prove by induction that $3^{4n + 2} + 1$ is divisible by $10$ Prove by induction: $3^{(4n+2)} + 1$ is divisible by $10$.
My basic step: $3^{(4n+2)} + 1$, where $n = 1$ gives me $3^6 + 1 = 730$, which is divisible by $10$. However, then I have to do the induction hypothesis and I am kind of stuck because I do not have an equality. How do I finish proving this by induction?
Many thanks.
Edit: I am thinking of creating a formula which involves $10n$? Would this be correct?

Suppose $3^{4k+2} +1$ is divided by 10, then we need to show $3^{4(k+1)+2} +1$ is also divided by 10. Note that:  
\begin{align*}
\ 3^{4(k+1)+2} +1 &= 3^{4k+2+4}+(81-80)
\\ &= 81\cdot3^{4k+2} + 81 - 80
\\ &= 81\cdot(3^{4k+2}+1) - 80 
\\ &= 81\cdot10m -80\ldots\ldots (\text{where}~3^{4k+2}+1 = 10m ~~\text{for some integer}~m)
\\ &= 10(81m-8)
\end{align*}
Hence, $3^{4(k+1)+2} +1$ is also divided by 10, which complete the proof.
