CDF for Negative Binomial Distribution I am trying to show that the following statement is true.
$$
\sum_{x = r}^{X}\binom{x-1}{r-1}p^r(1-p)^{x-r} =
\sum_{x = r}^{X}\binom{X}{x}p^x(1-p)^{X-x}
$$
Where $X$ and $r$ and $p$ are constants, with $X \geq r$, and $ 0 \leq p \leq 1.$
How did I get there? Well, this is the story:
Consider a sequence of independent binomial trials, each one producing the result success or failure, with probabilities $p$, and $1-p$, respectively.
Let $x$ be the total number of trials which must be carried out in order to attain exactly $r$ successes.
Knowing that the probability mass function for this Negative Binomial Distribution is as follows,
$P(x=X)=\binom{X-1}{r-1}p^r(1-p)^{X-r}$, (for $X \geq r$),
I was trying to prove the following about the corresponding Cumulative Distribution Function.
$P(x \leq X)=\sum_{x=r}^{x=X}\binom{x-1}{r-1}p^r(1-p)^{x-r}=1-\sum_{x=0}^{x=r-1}\binom{X}{x}p^x(1-p)^{X-x}$     
I started out with the following:
$\sum_{x=r}^{\infty}\binom{x-1}{r-1}p^r(1-p)^{x-r}=1$, 
which can be recast as below. (Relation I)
$\sum_{x=r}^{x=X}\binom{x-1}{r-1}p^r(1-p)^{x-r}+\sum_{x=X+1}^{\infty}\binom{x-1}{r-1}p^r(1-p)^{x-r}=1$ 
In addition, from binomial theorem, we have:
$\left ( p+(1-p) \right )^X=\sum_{x=0}^{x=X}\binom{X}{x}p^x(1-p)^{X-x}=1$
Which can be restated in Relation II as below.
$\sum_{x=0}^{x=r-1}\binom{X}{x}p^x(1-p)^{X-x}+\sum_{x=r}^{x=X}\binom{X}{x}p^x(1-p)^{X-x}=1$
Comparing the relations I and II with the expression for the CDF, the proof boils down to verification of the following:
$\sum_{x=r}^{x=X}\binom{x-1}{r-1}p^r(1-p)^{x-r}=\sum_{x=r}^{x=X}\binom{X}{x}p^x(1-p)^{X-x}$
Any idea how to continue form this point onward?
Thanks.

Suppose we seek to prove that
$$\sum_{k=0}^m {n+k-1\choose k} q^k
= \sum_{k=0}^m {m+n\choose m-k} q^{m-k} (1-q)^k.$$
The RHS is
$$\sum_{k=0}^m {m+n\choose k} q^{k} (1-q)^{m-k}.$$
Extracting the coefficient on $q$ on the RHS we get
$$[q^p] \sum_{k=0}^m {m+n\choose k} q^{k} (1-q)^{m-k}
= \sum_{k=0}^m {m+n\choose k} [q^p] q^{k} (1-q)^{m-k}
\\ = \sum_{k=0}^m {m+n\choose k} [q^{p-k}] (1-q)^{m-k}.$$
Now
$$[q^{p-k}] (1-q)^{m-k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{p-k+1}} (1-z)^{m-k} \; dz.$$
Observe that this vanishes when $p\lt  k$ as needed. Now by inspection
of the original RHS we see that we may assume $p\le m.$ Therefore when
$k\gt m$ the integral vanishes and we  may extend the range of the sum
to $m+n$, getting
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{p+1}} (1-z)^{m} 
\sum_{k=0}^{m+n} {m+n\choose k} \frac{z^k}{(1-z)^k}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{p+1}} (1-z)^{m} 
\left(1+\frac{z}{1-z}\right)^{m+n}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{p+1}} (1-z)^{m} 
\frac{1}{(1-z)^{m+n}}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{p+1}}
\frac{1}{(1-z)^{n}}
\; dz
\\ = {p+n-1\choose p}.$$
This is the claim.
