In the middle of a problem, I faced to the following series, \begin{eqnarray} \sum_{i=0}^{j-1} {j-i+k-1 \choose k-1} \left(\frac{a}{x+y}\right)^i \left(\frac{y-a}{x+y}\right)^{j-i} \end{eqnarray} I thought I could use Binomial theorem to find a closed form for it, but because of ${j-i+k-1 \choose k-1}$, I could not do it. Anyone could give me a clue that whether I can go further or not and how?
2026-04-07 06:28:59.1775543339
Obtaining a closed form for a series
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