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. 
 A: You can use the Hockey-stick identity. Using $\binom{i}{j}+\binom{i}{j+1} =\binom{i+1}{j+1}$ recursively, we have
$$\begin{align*}
\sum_{k=1}^n {2n-k \choose n} &= \color{red}{\binom{n}{n}}+\binom{n+1}{n}+\cdots+\binom{2n-1}{n}\\
&=\color{red}{\binom{n+1}{n+1}}+\binom{n+1}{n}+\cdots+\binom{2n-1}{n}
\\&=\binom{n+2}{n+1}+\binom{n+2}{n}+\cdots+\binom{2n-1}{n}
\\&=\binom{n+3}{n+1}+\binom{n+3}{n}+\cdots+\binom{2n-1}{n}
\\&=\cdots\\&=\binom{2n}{n+1} =\frac{2n\binom{2n-1}{n}}{n+1}.
\end{align*}$$
A: $\displaystyle \sum_{k=1}^{n}\binom{2n-k}{n}$ is the coefficient of $x^n$ in the expansion of $\displaystyle \sum_{k=1}^n(1+x)^{2n-k}$.
Note that $\displaystyle \sum_{k=1}^n(1+x)^{2n-k}=\frac{(1+x)^n[(1+x)^n-1]}{(1+x)-1}=\frac{(1+x)^{2n}-(1+x)^n}{x}$.
Therefore, $\displaystyle \sum_{k=1}^{n}\binom{2n-k}{n}=\binom{2n}{n+1}=\frac{(2n)!}{(n+1)!(n-1)!}=\frac{2n}{n+1}\binom{2n-1}{n}$.
A: By substituting $k\mapsto n-k$, we get
$$
\sum_{k=1}^n\binom{2n-k}{n}=\sum_{k=0}^{n-1}\binom{n+k}{n}\tag1
$$
We can derive a closed form for a generalization of the sum on the right hand side of $(1)$ using the expansion
$$
\sum_{k=0}^\infty\binom{n+k}{n}x^k=(1-x)^{-n-1}\tag2
$$
Taking the product of $(2)$ for two different exponents, we get
$$
\begin{align}
(1-x)^{-n-1}(1-x)^{-m-1}
&=\sum_{k=0}^\infty\binom{n+k}{n}x^k\sum_{j=0}^\infty\binom{m+j}{m}x^j\tag3\\
&=\sum_{k=0}^\infty\sum_{j=0}^k\binom{n+k-j}{n}\binom{m+j}{m}x^k\tag4\\
(1-x)^{-n-m-2}
&=\sum_{k=0}^\infty\binom{n+m+k+1}{n+m+1}x^k\tag5
\end{align}
$$
Equating the coefficients of $x^k$ in $(3)$ and $(4)$, we get
$$
\bbox[5px,border:2px solid #C0A000]{\sum_{j=0}^k\binom{n+k-j}{n}\binom{m+j}{m}=\binom{n+m+k+1}{n+m+1}}\tag6
$$
Substituting
$$
\begin{matrix}
n&\to&0\\
m&\to&n\\
k&\to&n-1\\
j&\to&k\\
\end{matrix}\tag7
$$
into $(6)$ yields
$$
\sum_{k=0}^{n-1}\binom{0+n-1-k}{0}\binom{n+k}{n}=\binom{0+n+n}{0+n+1}\tag8
$$
which evaluates to
$$
\sum_{k=0}^{n-1}\binom{n+k}{n}=\binom{2n}{n+1}\tag9
$$
Applying $(9)$ to $(1)$ yields
$$
\begin{align}
\sum_{k=1}^n\binom{2n-k}{n}
&=\binom{2n}{n+1}\tag{10}\\[6pt]
&=\frac{2n}{n+1}\binom{2n-1}{n}\tag{11}
\end{align}
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mc}[1]{\mathcal{#1}}
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$\ds{\sum_{k = 1}^{n}{2n - k \choose n} =
{2n{2n - 1 \choose n} \over n + 1}:\ {\LARGE ?}}$.

\begin{align}
&\bbox[10px,#ffd]{\sum_{k = 1}^{n}{2n - k \choose n}} =
\sum_{k = 1}^{\infty}\overbrace{2n - k \choose n - k}^{\ds{Symmetry}}
\ =\
\sum_{k = 1}^{\infty}\overbrace{{-n - 1 \choose n - k}
\pars{-1}^{n - k}}^{\ds{Negation}}
\\[5mm]= &\
\sum_{k = 1}^{\infty}
\pars{-1}^{n - k}\bracks{z^{n - k}}\pars{1 + z}^{-n - 1} =
\pars{-1}^{n}\bracks{z^{n}}\pars{1 + z}^{-n - 1}
\sum_{k = 1}^{\infty}\pars{-z}^{k}
\\[5mm] = &\
\pars{-1}^{n}\bracks{z^{n}}\pars{1 + z}^{-n - 1}\,
{-z \over 1 - \pars{-z}} =
\pars{-1}^{n + 1}\bracks{z^{n - 1}}\pars{1 + z}^{-n - 2}
\\[5mm] & =
\pars{-1}^{n + 1}{-n - 2 \choose n - 1} =
\pars{-1}^{n + 1}{2n \choose n - 1}\pars{-1}^{n - 1} =
\bbx{2n \choose n - 1}
\\[5mm] = &\
{\pars{2n}\pars{2n - 1}! \over \pars{n - 1}!\pars{n + 1}n!} =
\bbx{{2n{2n - 1 \choose n} \over n + 1}}
\end{align}
