combinatorics - Prove $sum_{i=0}^{i=x} {x choose i} {y+i choose x}+sum_{i=0}^{i=x} {x choose i} {y+1+i choose x}$

How to prove that
$$\sum_{i=0}^{i=x} {x \choose i} {y+i \choose x}+\sum_{i=0}^{i=x} {x \choose i} {y+1+i \choose x}=\sum_{i=0}^{i=x+1} {x+1 \choose i} {y+i \choose x}$$ ?

I tried to break the right side of equation down:

$$\sum_{i=0}^{i=x+1} {x+1 \choose i} {y+i \choose x}=\sum_{i=0}^{i=x} {x+1 \choose i} {y+i \choose x}+{x+1 \choose x+1} {y+x+1 \choose x}$$

Then I tried Vandermonde's Identity:
$${y+x+1 \choose x} = \sum_{i=0}^{i=x} {y+1 \choose i}{x \choose x-i}$$

Now I am totally lost. Can someone please tell me how to prove this equation?

Answer

This is very easy to prove using the integral representation of
binomial coefficients. Suppose we are trying to prove that

$$\sum_{k=0}^n {n\choose k} {m+k\choose n}
+ \sum_{k=0}^n {n\choose k} {m+1+k\choose n}
= \sum_{k=0}^{n+1} {n+1\choose k} {m+k\choose n}.$$

Use the two integrals
$${m+k\choose n}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{m+k}}{z^{n+1}} \; dz$$
and
$${m+k+1\choose n}

= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{m+1+k}}{z^{n+1}} \; dz.$$

This yields for the LHS the following sum consisting of two terms:
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^m}{z^{n+1}}
\sum_{k=0}^n {n\choose k} (1+z)^k\; dz
\\ + \frac{1}{2\pi i}
\int_{|z|=\epsilon}

\frac{(1+z)^{m+1}}{z^{n+1}}
\sum_{k=0}^n {n\choose k} (1+z)^k\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^m}{z^{n+1}} (2+z)^n \; dz
+ \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{m+1}}{z^{n+1}} (2+z)^n \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}

\frac{(1+z)^m}{z^{n+1}} (2+z)^{n+1} \; dz.$$

For the RHS we get the integral
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^m}{z^{n+1}}
\sum_{k=0}^{n+1} {n+1\choose k} (1+z)^k\; dz
= \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^m}{z^{n+1}} (2+z)^{n+1} \; dz.$$

The integrals on the LHS and on the RHS are the identical,
QED.

A similar calculation is at this
MSE link.

A trace as to when this method appeared on MSE and by whom starts at this

MSE link.