The problome is rewriten here:
$\sum_{i=0}^k \frac{((i-k)a)^i b^{k-i}}{i!}$
where $0
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!
Can we simplify the sum $\sum_{i=0}^k \frac{((i-k)a)^i b^{k-i}}{i!}$?
2
$\begingroup$
factorial
summation
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0What is the question? How do you want to simplify this? – 2012-10-29
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0It is equivalent to $\sum_{i=0}^k\prod_{j=1}^i (1-\frac{k}{i})a$. Is this helpful? – 2012-10-29
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0I want to get an expression without factorials or \sum operations. – 2012-10-29
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0Does [] mean floor or brackets? – 2012-10-29
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0They are only brackets. I have replaced [] with (). – 2012-10-29
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0Since it is a finite sum, try writing the first few terms and the last few terms, perhaps there is some simplification. – 2012-10-29