Is there an algebraic proof for $\sum_{m=k}^{n-k} \binom{m}{k}\binom{n-m}{k} = \binom{n+1}{2k+1}, n\ge2k\ge0$ $\sum_{m=k}^{n-k} \binom{m}{k}\binom{n-m}{k} = \binom{n+1}{2k+1}, n\ge2k\ge0$
An combinatorial proof of the identity above states as follow:
(1)Number of ways of picking (2k+1) numbers from 1 to (n+1) should be $\binom{n+1}{2k+1}$ 
(2)We pick (2k+1) numbers from 1 to (n+1) with median value (m+1).  Then, k numbers must be selected from 1~m, and the other k numbers must be chosen from (m+2)~(n+1). Thus there are $\binom{m}{k}\binom{n-m}{k}$ ways for picking (2k+1) numbers with median value (m+1). Since $n-k\ge m\ge k$, there are total $\sum_{m=k}^{n-k} \binom{m}{k}\binom{n-m}{k}$ ways.
Since (1)=(2), the statement is true.
But is it possible to sketch an algebraic proof that doesn't require building combinatorial models? 

Here is an algebraic proof based upon generating functions. It is convenient  to use the coefficient of operator  $[z^n]$ to  denote the coefficient of  $z^n$  in  a series. This way we can write for instance
\begin{align*}
[z^k](1+z)^n=\binom{n}{k}
\end{align*}

We obtain
  \begin{align*}
\color{blue}{\sum_{m=k}^{n-k}}&\color{blue}{\binom{m}{k}\binom{n-m}{k}}\\
&=\sum_{m=0}^{n-2k}\binom{m+k}{m}\binom{n-m-k}{k}\tag{1}\\
&=\sum_{m=0}^{n-2k}\binom{-k-1}{m}(-1)^m\binom{n-m-k}{k}\tag{2}\\
&=\sum_{m=0}^\infty[z^m](1-z)^{-k-1}[u^k](1+u)^{n-m-k}\tag{3}\\
&=[u^k](1+u)^{n-k}\sum_{m=0}^\infty\left(\frac{1}{1+u}\right)^{m}[z^m](1-z)^{-k-1}\tag{4}\\
&=[u^k](1+u)^{n-k}\left(1-\frac{1}{1+u}\right)^{-k-1}\tag{5}\\
&=[u^k](1+u)^{n-k}u^{-k-1}(1+u)^{k+1}\tag{6}\\
&=[u^{2k+1}](1+u)^{n+1}\tag{7}\\
&\color{blue}{=\binom{n+1}{2k+1}}\tag{8}
\end{align*}
  and the claim follows.

Comment:


*

*In (1) we shift the index to start with $m=0$ and use the binomial identity $\binom{p}{q}=\binom{p}{p-q}$.

*In (2) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

*In (3) we apply the  coefficient of operator twice and we set the upper bound of the series to $\infty$ without changing anything since we are adding zeros only.

*In (4) we  use the linearity of the coefficient of operator and we apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.

*In (5) we apply the substitution rule of the coefficient of operator with $z=\frac{1}{1+u}$
\begin{align*}
A(u)=\sum_{m=0}^\infty a_m u^m=\sum_{m=0}^\infty u^m [z^m]A(z)
\end{align*}

*In (6) we do some simplifications.

*In (7) we do some more simplifications and apply the same rule as we did in (4).

*In (8) we select the coefficient of $u^{2k+1}$.
