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I got something like

$\displaystyle\sum_{i=0}^K{ \binom{n+i}{i} \cdot \alpha^i} $

where $n,\ K,\ \alpha$ are definite values, $\binom{n+i}{i}$ is the Combinatorial number that choose $i$ from $n+i$, can this summation be simplified? Thank you.

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  • $\begingroup$ For $k\to\infty$ we have $S=\dfrac1{(1-\alpha)^{n+1}}.~$ See binomial series. $\endgroup$ – Lucian Dec 19 '14 at 3:47
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The summand, being a product of a binomial coefficient and something to the power of the iterand, looks like a hypergeometric function.

In fact, symbolic algebra reveals that it is equal to $$ \left(1-\alpha\right)^{-n-1}-\alpha^{K+1}\binom{K+n+1}{K+1}{}_{2}F_{1}\left(1,K+n+2;K+2;\alpha\right). $$ Naturally, the trouble term is the hypergeometric $_{2}F_{1}$. I do not think this is a particularly useful form, other than being "closed" in some sense.

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  • $\begingroup$ Thank you very much, actually I want to simplify $\frac{\binom{n+K}{K} \alpha^K } {\sum_{i=0}^K{ \binom{n+i}{i} \cdot \alpha^i}} $, is it possible to remove the hypergeometric function? can this be simplified to some closed form? $\endgroup$ – JLiu Dec 19 '14 at 5:07
  • $\begingroup$ Not as far as I know. Your expression would be equivalent to $$ \frac{\alpha^{K}\left(1-\alpha\right)^{n+1}\binom{K+n}{K}}{1-\alpha^{K+1}\left(1-\alpha\right)^{n+1}\binom{K+n+1}{K+1}{}_{2}F_{1}\left(1,K+n+2;K+2;\alpha\right)}. $$ $\endgroup$ – parsiad Dec 19 '14 at 7:32
  • $\begingroup$ Why do you care so much about "closed forms?" The distinction between something that is "closed" and not is arbitrary. All that matters is whether you can perform analysis on it or compute it quickly, depending on the situation. $\endgroup$ – parsiad Dec 19 '14 at 7:33
  • $\begingroup$ @JLiu Since we're dealing with integer parameters, we can use the fundamental recurrence relations to prove this identity, and so reduce the hypergeometric function to a finite binomial sum. $\endgroup$ – David H Dec 19 '14 at 8:44

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