# Evaluating (bounding) sum involving the binomial coefficient

Solving some problem I have stumbled into the following sum :

$\displaystyle \sum_{i=0}^{n-2} {e \choose i} (n-1-i) (1-p)^{e-i} p^i$

where $0 \leq e \leq {n \choose 2}$.

I am not very efficient with the evaluation of these sums so I would like to ask if there is any way to evaluate this sum or obtain a sharp lower bound for it?

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Did you get something out of a solution below? – Did Apr 7 '11 at 7:53

Consider $$S(k,e)=\sum_{i=0}^{k} {e \choose i} (1-p)^{e-i} p^i.$$ Then $S(k,e)=1$ if $k\ge e$, and your sum $S_{n,e}$ is $$S_{n,e}=(n-1)S(n-2,e)-epS(n-3,e-1).$$ Hence $S_{n,e}=n-1-ep$ for every $n\ge e+2$.
If $n<e+2$, there might not exist so simple expressions of $S_{n,e}$ in all generality. Note however for small $n$ that $S(0,e)=(1-p)^e$, hence $S_{0,e}=S_{1,e}=0$ and $S_{2,e}=(1-p)^e$.
Edit According to E.C. Molina (Application to the Binomial Summation of a Laplacian Method for the Evaluation of Definite Integrals, Bell System Technical Journal, v8: i1 January 1929, 99-108, available here), for $k<e$, $$S(k,e)=B_{k,e}(p)/B_{k,e}(0),\quad\mbox{where}\ B_{k,e}(p)=\int_p^1x^{k}(1-x)^{e-k-1}\mathrm{d}x.$$ The author then presents some approximations of $S(k,e)$.