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.

We may prove it this way too. (Symbols are a bit different, apologies for that).
We want to show that
$$
\sum_{j=r}^n \binom{j-1}{r-1}  (1-p)^{j-r}
 =  \sum_{j=r}^{n} \binom{n}{j} p^{j-r} (1-p)^{n-j}
$$
For $n=r$, LHS = 1 = RHS.
Assume the result is true for $n = r + m$, we prove it is true for $n = r + m + 1$. 
LHS =
$$ \sum_{j=r}^{r+m+1} \binom{j-1}{r-1}  (1-p)^{j-r}
 = \sum_{j=r}^{r+m} \binom{j-1}{r-1}  (1-p)^{j-r} +  \binom{r+m}{r-1} (1-p)^{m+1}\\
  =^{(Assumption)}  \sum_{j=r}^{r+m} \binom{r+m}{j} p^{j-r}  (1-p)^{r+m-j} +  \binom{r+m}{r-1} (1-p)^{m+1} \\
=(*) \sum_{j=r}^{r+m} \binom{r+m+1}{j} p^{j-r}  (1-p)^{r+m-j} - \sum_{j=r}^{r+m} \binom{r+m}{j-1} p^{j-r}  (1-p)^{r+m-j} + \binom{r+m}{r-1} (1-p)^{m+1} + p \sum_{j=r}^{r+m} \binom{r+m+1}{j} p^{j-r}  (1-p)^{r+m-j} - p\sum_{j=r}^{r+m} \binom{r+m+1}{j} p^{j-r}  (1-p)^{r+m-j} \\
= \sum_{j=r}^{r+m} \binom{r+m+1}{j} p^{j-r}  (1-p)^{r+m+1-j} + \sum_{j=r}^{r+m} \binom{r+m+1}{j} p^{1+j-r}  (1-p)^{r+m-j} - \sum_{j=r}^{r+m} \binom{r+m}{j-1} p^{j-r}  (1-p)^{r+m-j} + \binom{r+m}{r-1} (1-p)^{m+1} + p^{m+1} - p^{m+1}\\
= \sum_{j=r}^{r+m+1} \binom{r+m+1}{j} p^{j-r}  (1-p)^{r+m+1-j} + A.
$$
Here
$$A = (*) \sum_{j=r}^{r+m} \Bigg\{ \binom{r+m}{j} + \binom{r+m}{j-1}  \Bigg\} p^{1+j-r}  (1-p)^{r+m-j} - \sum_{j=r}^{r+m} \binom{r+m}{j-1} p^{j-r}  (1-p)^{r+m-j} +  \binom{r+m}{r-1}  (1-p)^{m+1} - p^{m+1}\\
 = \sum_{j=r}^{r+m} \binom{r+m}{j} p^{1+j-r}  (1-p)^{r+m-j} - \sum_{j=r}^{r+m} \binom{r+m}{j-1} p^{j-r}  (1-p)^{r+m+1-j} + \binom{r+m}{r-1}  (1-p)^{m+1} - p^{m+1}\\
= \sum_{j=r}^{r+m} \binom{r+m}{j} p^{1+j-r}  (1-p)^{r+m-j}  - \sum_{j'=r-1}^{r+m-1} \binom{r+m}{j'} p^{j'-r+1}  (1-p)^{r+m-j'} + \binom{r+m}{r-1}  (1-p)^{m+1} - p^{m+1} = 0.
 $$
The equalities with $(*)$ make use of [Pascal rule]https://en.wikipedia.org/wiki/Pascal%27s_rule#:~:text=In%20mathematics%2C%20Pascal's%20rule%20(or,natural%20numbers%20n%20and%20k%2C&text=is%20a%20binomial%20coefficient%3B%20one,(1%20%2B%20x)n.
