Sum over two binomials identity So while trying to count the number of configurations in a statistical mechanics research problem I come across this lovely sum:
$$\sum_{i=0}^k \binom{i+r}{r} \binom{k-i+r}{r}$$
I scoured the internet for binomial identities the closest being an identity close to the Chu-Vandermonde, so after a few hours I give up and chuck it into Mathematica who immediately gives me:
$$\sum_{i=0}^k \binom{i+r}{r} \binom{k-i+r}{r}=\binom{k+2r+1}{k}$$
Was just wondering if anybody had any idea how to show that or if it is a standard identity I just don't know.
BONUSES:
Mathematica wasn't able to do it, but if anybody has any idea or insight on how to do a double sum over three binomials:
$$ \sum_{i=0}^N \sum_{j=0}^{N-i} \binom{i+r}{r}\binom{j+r}{r}\binom{N-i-j+r}{r}$$
It would help gain insight on trying to do it over K binomials:
$$\sum_{x_1, \cdots, x_K = 0 }^{N} \  \prod_{i=1}^K  \binom{x_i+r}{r}  \delta^{N}_{\sum_{i=1}^K x_i}$$
where $\delta_i^j$ is the Kronecker delta.
 A: Using a combination of upper negation and Vandermonde, we have:
$$\begin{align}\sum_{i=0}^k \binom{i+r}{r} \binom{k-i+r}{r}
&=\sum_{i=0}^k\binom {i+r}i\binom {k-i+r}{k-i}\\
&=\sum_{i=0}^k (-1)^i\binom{-r-1}i (-1)^{k-i}\binom {-r-1}{k-i}\\
&=(-1)^k\sum_{i=0}^k\binom {-r-1}i\binom{-r-1}{k-i}\\
&=(-1)^k\binom{-2r-2}k\\
&=(-1)^{2k}\binom{k+2r+1}k\\
&=\binom{k+2r+1}{k}\qquad\blacksquare
\end{align}$$

We can use the result above to help with the bonus case.
$$\begin{align}
\sum_{i=0}^N \sum_{j=0}^{N-i} \binom{i+r}{r}\binom{j+r}{r}\binom{N-i-j+r}{r}
&=\sum_{i=0}^N \binom{i+r}{r}\color{blue}{\sum_{j=0}^{N-i}\binom{j+r}{r}\binom{\overline{N-i}-j+r}{r}}\\
&=\sum_{i=0}^N \binom{i+r}{r}\color{blue}{\binom{\overline{N-i}+2r+1}{\overline{N-i}}}\\
&=\sum_{i=0}^N (-1)^i\binom{-r-1}{i}(-1)^{N-i}\binom{-2r-2}{N-i}\\
&=(-1)^N\sum_{i=0}^N \binom{-r-1}i\binom{-2r-2}{N-i}\\
&=(-1)^N\binom{-2r-3}N\\
&=\binom{N+2r+2}N\qquad\blacksquare
\end{align}$$
An interesting pattern emerges...
A: Suppose we  seek to evaluate
$$S(k,r) = \sum_{q=0}^k {q+r\choose r} {k-q+r\choose r}
= \sum_{q=0}^k {q+r\choose r} {k-q+r\choose k-q}.$$
Introduce
$${k-q+r\choose k-q} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{k-q+1}} (1+z)^{k-q+r} \; dz.$$
Observe that when  $q\gt k$ the integral vanishes so we  may use it to
control the range, extending the sum in $q$ to infinity to get
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{k+1}} (1+z)^{k+r}
\sum_{q\ge 0} {q+r\choose r} \frac{z^q}{(1+z)^q} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{k+1}} (1+z)^{k+r}
\frac{1}{(1-z/(1+z))^{r+1}} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{k+1}} (1+z)^{k+2r+1}
\frac{1}{(1+z-z)^{r+1}} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{k+1}} (1+z)^{k+2r+1} \; dz
\\ = {k+2r+1\choose k}.$$
The general pattern here for $K$ binomials is
$$[z^N] \frac{1}{(1-z)^{(r+1)\times K}}
= {(r+1)K - 1+ N\choose N}.$$
