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Topic: SUM
Replies: 8   Last Post: Feb 11, 1999 3:12 AM

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Stephen Montgomery-Smith

Posts: 97
Registered: 12/6/04
Posted: Feb 10, 1999 10:48 AM
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Clive Tooth wrote:
> Franck Dewannieux wrote:

> > Could anyone calculate
> > Summ(k=0, k=n) {Cnk.k^2 .p^k.p^(n-k)} with Cnk = n!/(k!(n-k)!)

> Do you mean p^n.Sum(k=0, k=n){Cnk.k^2} ?

I think he/she meant:

Summ(k=0, k=n) {Cnk.k^2 .p^k.(1-p)^(n-k)} with Cnk = n!/(k!(n-k)!)

I get (using Mathematica)

n p n
n (1 - p) p (1 - p + n p) (1 + -----)
1 - p


Stephen Montgomery-Smith stephen@math.missouri.edu
307 Math Science Building stephen@showme.missouri.edu
Department of Mathematics stephen@missouri.edu
University of Missouri-Columbia
Columbia, MO 65211

Phone (573) 882 4540
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