Re: Trying to solve a sum
- To: mathgroup at smc.vnet.net
- Subject: [mg36088] Re: [mg36070] Trying to solve a sum
- From: Rob Pratt <rpratt at email.unc.edu>
- Date: Wed, 21 Aug 2002 05:51:53 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
On Thu, 15 Aug 2002, Constantine wrote:
> 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:
[messy expression involving Hypergeometric2F1 snipped]
Since (1 + n - 2 k) / (1 + n - k) = 1 - k / (n - k + 1), we have
Binomial[n, k] * (1 + n - 2 k) / (1 + n - k)
= Binomial[n, k] * (1 - k / (1 + n - k))
= Binomial[n, k] - Binomial[n, k] * k / (1 + n - k))
= Binomial[n, k] - n! * k / (k! (n - k)! (1 + n - k))
= Binomial[n, k] - n! / ((k - 1)! (n - k + 1)!)
= Binomial[n, k] - Binomial[n, k - 1].
Now if the p's weren't there (or p = 1 - p, equivalently p = 1 / 2), we
would have a telescoping sum. With general p, the following is the best I
can do.
Sum[ p^(n-k) (1-p)^k Binomial[n, k] * (1+ n -2 k) / (1+n-k), {k, 0,
Ceiling[n/2]} ]
= Sum[ p^(n-k) (1-p)^k Binomial[n, k], {k, 0, Ceiling[n/2]} ]
- Sum[ p^(n-k) (1-p)^k Binomial[n, k - 1], {k, 1, Ceiling[n/2]} ]
= Sum[ p^(n-k) (1-p)^k Binomial[n, k], {k, 0, Ceiling[n/2]} ]
- Sum[ p^(n-(j+1)) (1-p)^(j+1) Binomial[n, j], {j, 0, Ceiling[n/2] - 1} ]
= Sum[ p^(n-k) (1-p)^k Binomial[n, k], {k, 0, Ceiling[n/2] - 1} ]
+ p^(n-Ceiling[n/2]) (1-p)^Ceiling[n/2] Binomial[n, Ceiling[n/2]]
- ((1 - p) / p) Sum[ p^(n-j) (1-p)^j Binomial[n, j], {j, 0, Ceiling[n/2] - 1} ]
= (2 - 1 / p) Sum[ p^(n-k) (1-p)^k Binomial[n, k], {k, 0, Ceiling[n/2] - 1} ]
+ p^(n-Ceiling[n/2]) (1-p)^Ceiling[n/2] Binomial[n, Ceiling[n/2]]
Note that if p = 1 / 2, the first line disappears, leaving us with
Binomial[n, Ceiling[n/2]] / 2^n.
I don't believe the partial binomial sum has a simple form.
Rob Pratt
Department of Operations Research
The University of North Carolina at Chapel Hill
rpratt at email.unc.edu
http://www.unc.edu/~rpratt/