Re: Trying to solve a sum
- To: mathgroup at smc.vnet.net
- Subject: [mg36080] Re: Trying to solve a sum
- From: "Borut L" <gollum at email.si>
- Date: Wed, 21 Aug 2002 05:51:44 -0400 (EDT)
- References: <ajfini$ipa$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Hi, I don't think there is any simpler analytical solution. Though, if you let n be large, the sum can be well aproximated by the gauss integral, i.e. the error function. bye, Borut "Constantine" <celster at cs.technion.ac.il> wrote in message news:ajfini$ipa$1 at smc.vnet.net... > Hi, > I'm trying to solve the following sum: > > Sum[ p^(n-k) (1-p)^k Binomial[n, k] * (1+ n -2 k) / (1+n-k), {k, 0, > Ceiling[n/2]} ] > > Pls, if anyone knows if that sum has a simple solution, I'll be pleasant > for a hint how to find it. > > Thanks in advance. > Constantine. > > > > P.S. The Mathematica produces the following output for this sum: > > > n 1 1 + n > Out[1]= p ((-) (-1 + 2 p) - > p > > 1 Ceiling[n/2] > > ((-1 + -) (-1 + p) Gamma[1 + n] > p > > n > > ((1 + n) Gamma[n - Ceiling[-]] > 2 > > n n -1 + p > > Hypergeometric2F1[1, -n + Ceiling[-], 2 + Ceiling[-], ------] - > 2 2 p > > n > > 2 Gamma[1 + n - Ceiling[-]] > 2 > > n n -1 + p > > Hypergeometric2F1[1, 1 - n + Ceiling[-], 2 + Ceiling[-], ------]) > 2 2 p > > n n > > ) / (p Gamma[n - Ceiling[-]] Gamma[1 + n - Ceiling[-]] > 2 2 > > n > > Gamma[2 + Ceiling[-]])) > 2 > > > > > > > > Constantine Elster > Computer Science Dept. > Technion I.I.T. > Office: Taub 411 > Tel: +972 4 8294375 >