How prove this$\sum_{i=0}^{m-1}\binom{n-1+i}{i}x^ny^i+\sum_{j=0}^{n-1}\binom{m-1+j}{j}x^my^j=1$ let $m,n$ be positive numbers,and $x,y>0$ such $x+y=1$,show that
$$\sum_{i=0}^{m-1}\binom{n-1+i}{i}x^ny^i+\sum_{j=0}^{n-1}\binom{m-1+j}{j}x^jy^m=1$$
My try:
$$\sum_{i=0}^{m-1}\binom{n-1+i}{i}x^ny^i=\sum_{i=0}^{m-1}\binom{n-1+i}{i}x^n(1-x)^i=\sum_{i=0}^{m-1}\binom{n-1+i}{i}x^n\sum_{k=0}^{i}(-1)^k\binom{i}{k}x^k$$

Suppose we seek to show that
$$\sum_{q=0}^{m-1} {n-1+q\choose q} x^n (1-x)^q
+ \sum_{q=0}^{n-1} {m-1+q\choose q} x^q (1-x)^m = 1$$
where $n,m\ge 1.$
We will evaluate  the second term by a contour  integral and show that
is equal to one minus the first term which is the desired result.
Introduce the Iverson bracket
$$[[0\le q\le n-1]]
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{z^q}{z^n} 
\frac{1}{1-z} \; dz.$$
With this bracket we may extend the sum in $q$ to infinity to get
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^n} 
\frac{1}{1-z} 
\sum_{q\ge 0} {m-1+q\choose q} z^q x^q (1-x)^m\; dz
\\ = \frac{(1-x)^m}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^n} 
\frac{1}{1-z}
\sum_{q\ge 0} {m-1+q\choose q} z^q x^q \; dz
\\ = \frac{(1-x)^m}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^n} 
\frac{1}{1-z}
\frac{1}{(1-xz)^m} \; dz.$$
Now we  have three poles here at  $z=0$ and $z=1$ and  $z=1/x$ and the
residues at these  poles sum to zero, so we  can evaluate the residue
at  zero  by computing  the  negative of  the  residues  at $z=1$  and
$z=1/x.$
Observe that the  residue at infinity is zero as can  be seen from the
following computation:
$$-\mathrm{Res}_{z=0} \frac{1}{z^2}
z^n \frac{1}{1-1/z}\frac{1}{(1-x/z)^m}
\\ -\mathrm{Res}_{z=0} \frac{1}{z^2}
z^n \frac{z}{z-1}\frac{z^m}{(z-x)^m}
\\ -\mathrm{Res}_{z=0}
z^{n+m-1} \frac{1}{z-1}\frac{1}{(z-x)^m} = 0.$$
Returning to the main thread the residue at $z=1$ as seen from
$$- \frac{(1-x)^m}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^n} 
\frac{1}{z-1}
\frac{1}{(1-xz)^m} \; dz.$$
is $$-(1-x)^m \frac{1}{(1-x)^m} = -1.$$
For the residue at $z=1/x$ we consider
$$\frac{(1-x)^m}{x^m \times 2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^n} 
\frac{1}{1-z}
\frac{1}{(1/x-z)^m} \; dz
\\ = \frac{(-1)^m (1-x)^m}{x^m \times 2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^n} 
\frac{1}{1-z}
\frac{1}{(z-1/x)^m} \; dz.$$
and use the following derivative:
$$\frac{1}{(m-1)!}
\left(\frac{1}{z^n} \frac{1}{1-z}\right)^{(m-1)}
\\ = \frac{1}{(m-1)!}
\sum_{q=0}^{m-1} {m-1\choose q}
\frac{(-1)^q (n+q-1)!}{(n-1)! z^{n+q}}
\frac{(m-1-q)!}{(1-z)^{m-q}}
\\ = 
\sum_{q=0}^{m-1} \frac{1}{q!}
\frac{(-1)^q (n+q-1)!}{(n-1)! z^{n+q}}
\frac{1}{(1-z)^{m-q}}
\\ = 
\sum_{q=0}^{m-1} {n+q-1\choose q}
\frac{(-1)^q}{z^{n+q}}
\frac{1}{(1-z)^{m-q}}.$$
Evaluate this at $z=1/x$ and multiply by the factor in front to get
$$\frac{(-1)^m (1-x)^m}{x^m} \times
\sum_{q=0}^{m-1} {n+q-1\choose q}
(-1)^q x^{n+q}
\frac{1}{(1-1/x)^{m-q}}
\\ = \frac{(-1)^m (1-x)^m}{x^m} \times
\sum_{q=0}^{m-1} {n+q-1\choose q}
(-1)^q x^{n+q}
\frac{x^{m-q}}{(x-1)^{m-q}}
\\ = (-1)^m (1-x)^m \times
\sum_{q=0}^{m-1} {n+q-1\choose q}
(-1)^q x^{n} (-1)^{m-q}
\frac{1}{(1-x)^{m-q}}
\\ = 
\sum_{q=0}^{m-1} {n+q-1\choose q}
x^{n} (1-x)^q.$$
This yields for the second sum term the value
$$1 - \sum_{q=0}^{m-1} {n+q-1\choose q}
x^{n} (1-x)^q$$
showing that when we add the  first and the second sum by cancellation
the end result is one, as claimed.
This identity generalizes an identity by Gosper to be found at this MSE link.
Remark Fri Jun 9 2017. As written this proof requires $|1/x| \gt \epsilon$ or $1/\epsilon \gt |x|$ for the pole at $1/x$ to be outside the circular contour and for the geometric series to converge. Note however that these are polynomials in $x$ of degree $n+m-1$ and hence this is sufficient to show they agree for all $x.$  
