# Can we simplify the sum $\sum_{i=0}^k \frac{((i-k)a)^i b^{k-i}}{i!}$?

The problome is rewriten here: $\sum_{i=0}^k \frac{((i-k)a)^i b^{k-i}}{i!}$ where $0<a<1$, k is an integer larger than 1.

I came to this equation when i try to find some probability. I have tried some formulas on permutation and combination, fractional, but with little improvement.

I hope you can give me some sugestions! Thanks a lot!

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What is the question? How do you want to simplify this? –  glebovg Oct 29 '12 at 3:05
It is equivalent to $\sum_{i=0}^k\prod_{j=1}^i (1-\frac{k}{i})a$. Is this helpful? –  Severals-user45972 Oct 29 '12 at 3:07
I want to get an expression without factorials or \sum operations. –  Severals-user45972 Oct 29 '12 at 3:09
Does [] mean floor or brackets? –  glebovg Oct 29 '12 at 3:13
They are only brackets. I have replaced [] with (). –  Severals-user45972 Oct 29 '12 at 3:14