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: $$\sum_{i=0}^{m-1}\binom{n-1+i}{i}x^ny^i=\sum_{i=0}^{m-1}\binom{n-1+i}{i}x^n(1-x)^i=\sum_{i=0}^{m-1}\binom{n-1+i}{i}x^n\sum_{k=0}^{i}(-1)^k\binom{i}{k}x^k$$

The exponents are reversed in the second summation: it should be

$$\sum_{j=0}^{n-1}\binom{n-1+j}jx^jy^m\;.$$

(You can see that the version in the problem is wrong by actual calculation with $m=n=2$ and $x=\frac14$, for instance.)

Think of $x$ as the probability of a success and $y$ as the probability of failure in a Bernoulli trial. The experiment consists of $m+n-1$ independent Bernoulli trials each with success probability $x$. Then

$$\binom{n-1+i}ix^ny^i$$

is the probability of getting the $n$-th success on the $(n+i)$-th trial (why?), so

$$\sum_{i=0}^{m-1}\binom{n-1+i}ix^ny^i$$

is the probability of getting at least $n$ successes. Similarly,

$$\sum_{j=0}^{n-1}\binom{n-1+j}jx^jy^m$$

is the probability of getting at least $m$ failures. To complete the proof, show that exactly one of these two events must occur.

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 $x.$