Proof of $\sum_{k=1}^n {2n-k \choose n} = \frac{2n{2n-1 \choose n}}{n+1} $ In a combinatorial math problem, I found the need to have a short form for $\sum_{k=1}^n {2n-k \choose n}$.
I searched it on Wolfram|Alpha and it gave me the result $\frac{2n{2n-1 \choose n}}{n+1}$ which indeed solves my problem, but no steps were available and I wasn't able to find a way to prove it by myself. I tried by writing the sum term by term and using ${n \choose k} = {n! \over k!(n-k)!}$ but it overcomplicated everything. Could anyone help me with this?
I'm sorry if this is a duplicate. As far as I searched, I haven't found any question regarding this sum, but maybe I haven't searched enough. 

You can use the Hockey-stick identity. Using $\binom{i}{j}+\binom{i}{j+1} =\binom{i+1}{j+1}$ recursively, we have
$$\begin{align*}
\sum_{k=1}^n {2n-k \choose n} &= \color{red}{\binom{n}{n}}+\binom{n+1}{n}+\cdots+\binom{2n-1}{n}\\
&=\color{red}{\binom{n+1}{n+1}}+\binom{n+1}{n}+\cdots+\binom{2n-1}{n}
\\&=\binom{n+2}{n+1}+\binom{n+2}{n}+\cdots+\binom{2n-1}{n}
\\&=\binom{n+3}{n+1}+\binom{n+3}{n}+\cdots+\binom{2n-1}{n}
\\&=\cdots\\&=\binom{2n}{n+1} =\frac{2n\binom{2n-1}{n}}{n+1}.
\end{align*}$$
