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?
 A: 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.

