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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?2011-04-07

1 Answers 1

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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

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

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    Didier, thanks for the link; I did not know until now that there is now a digital archive for *The Bell System Technical Journal*.2011-05-10
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    Yes. The treasure chest is here: http://www.alcatel-lucent.com/bstj/2011-05-10