Prove that $\sum_{k=0}^n \binom{3n-k}{2n}=\binom{3n+1}{n}$ 
Prove that $$\sum_{k=0}^n \binom{3n-k}{2n}=\binom{3n+1}{n}$$

I've tried multiple things that didn't work. 
Maybe this would help
$$\sum_{k=0}^n \binom{3n-k}{2n}=\sum_{k=0}^n \binom{3n-(n-k)}{2n}=\sum_{k=0}^n \binom{2n+k}{2n}$$

Your identity is a special case of the more general identity $$S(m,n) = \sum_{k=0}^n \binom{m+k}{m} = \binom{m+n+1}{m+1},$$ which you can prove by induction on $n$:  note $S(m,0) = 1 = \binom{m+1}{m+1}$.  Then observe $$\begin{align*} S(m,n+1) &= S(m,n) + \binom{m+n+1}{m} \\ 
&= \binom{m+n+1}{m+1} + \binom{m+n+1}{m} \\
&= \binom{m+n+2}{m+1} = \binom{m + (n+1) + 1}{m+1},
\end{align*}$$
hence by the induction hypothesis, the claim is proven.  Then choose $m = 2n.$
