# Can this binomial summation be simplified?

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.

• For $k\to\infty$ we have $S=\dfrac1{(1-\alpha)^{n+1}}.~$ See binomial series. – Lucian Dec 19 '14 at 3:47

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.
• 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? – JLiu Dec 19 '14 at 5:07
• Not as far as I know. Your expression would be equivalent to $$\frac{\alpha^{K}\left(1-\alpha\right)^{n+1}\binom{K+n}{K}}{1-\alpha^{K+1}\left(1-\alpha\right)^{n+1}\binom{K+n+1}{K+1}{}_{2}F_{1}\left(1,K+n+2;K+2;\alpha\right)}.$$ – parsiad Dec 19 '14 at 7:32