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Topic: Probability and Combinatorics
Replies: 0

 martin Posts: 2 Registered: 12/20/08
Probability and Combinatorics
Posted: Dec 20, 2008 8:04 PM

Hi!

Does anybody know how to solve the following problem:

$$\sum_{p=0}^{M}\binom{2M+2}{p}(M+1-p)=?$$

Well, actually i know that the solution is:

$$(2M+1)\binom{2M}{m}=\frac{2^{2M+1}\Gamma(\frac{3}{2}+M)}{\sqrt{\pi}\Gamma(M+1)}$$

but i cannot prove it (mathematica calculates the first sum and gives the answer).

Maybe someone knows the simple (combinatorial?) solution?

The original problem was:

Let $$R ^{M} _{P} = \sum_{s = 0}^{P} {M + 1 \choose s}$$, for $$0 \leqslant P \leqslant M$$, $$P,M\in \mathbb{N}$$

Proove that:
$$\sum_{q = 0}^{M}R^{M}_{q}\cdot R^{M}_{M - q} = (2M + 1) {2M \choose M}$$