I am trying to solve the sum of a specific type of binomial distribution:
\begin{align} \sum_{n=0}^{\kappa-s}\binom{s+n-1}{n}\left(1-x\right)^n \end{align}
The problem is that the sum affects the number of trials too so the binomial theorem does not work. I am not sure if I can use Stirling's formula as $n$ is bounded from above. The parameter of interest is $\kappa$, which I am using in an optimization problem. I looked on different posts here but none had the summation and the distribution included. So suggestions will be very helpful.
For $\kappa \ge s$ and $|1-x| < 1$ the sum can be written using a hypergeometric function as $$ \frac{1}{x^{s}}-{{\kappa}\choose{\kappa -s +1}} \left(1-x \right)^{\kappa -s +1} {}_{2}F_{1}^{}\! \left(1,\kappa +1;\kappa -s +2;1-x \right) $$ There is no more "elementary" form of this.
EDIT: This may be obtained as follows.
The sum of your series up to $\infty$ is
$$ \sum_{n=0}^\infty {s+n-1 \choose n} (1-x)^n = \dfrac{1}{x^s} \ {\rm if}\ |1-x| < 1$$ This is a binomial series. Now we need to subtract the sum from $\kappa-s+1$ to $\infty$.
Essentially by definition, $${}_2F_1(1,\kappa+1; \kappa-s+2; 1-x) = \frac{(\kappa-s+1)!}{\kappa!} \sum_{j=0}^\infty \dfrac{(\kappa+j)! }{(1+\kappa-s+j)! }(1-x)^j$$ Change index variable to $n = \kappa-s+1+j$ and we have
$$ {}_2F_1(1,\kappa+1; \kappa-s+2; 1-x) = \frac{(\kappa-s+1)!}{\kappa!} \sum_{n=\kappa-s+1}^\infty \dfrac{(n+s-1)!}{n! }(1-x)^{n-1-\kappa+s} $$