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$$ \sum\limits_{s=0}^{\infty}{p+s \choose s}{2p+m \choose 2p+2s} = 2^{m-1} \frac{2p+m}{m}{m+p-1 \choose p}$$

Class themes are: Generating functions and formal power series.

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looks like a combinatorial argument might be easier than doing just algebra. –  user17762 Nov 1 '11 at 20:11
2  
The sum is in fact from $0$ to $m/2$ since for $s>m/2$ you have ${2p+m \choose 2p+2s} =0$. –  Beni Bogosel Nov 1 '11 at 20:12
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If you want to do homework, you have to know what you have learned in your course. This is something we cannot do for you. Please say what you learned or I will just remark that this sum is utterly trivial using Zeilberger's algorithm. –  Phira Nov 1 '11 at 20:23
    
@BeniBogosel Who said that $m$ was an integer? –  Phira Nov 1 '11 at 20:24
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Who said that it isn't? –  Beni Bogosel Nov 1 '11 at 20:27

2 Answers 2

up vote 5 down vote accepted

Ok, here is an approach with generating functions. Let $$ g_1(z) = \sum_{s=0}^\infty \binom{p+s}{s} z^s = \frac{1}{\left(1-z\right)^{p+1}} $$

$$ g_2(z) = \sum_{s=0}^\infty \binom{2p+m}{s} z^s = \left(1+z\right)^{m+2p} $$ Now $$ \begin{eqnarray} \sum_{s=0}^\infty \binom{p+s}{s} \binom{2p+m}{2p+2s} &=& \sum_{s=0}^\infty \binom{p+s}{s} \binom{2p+m}{m-2s} = [z]^m g_1(z^2) g_2(z) = [z]^m \frac{\left(1+z\right)^{m+2p}}{(1-z^2)^{p+1}} \\ &=& [z]^m \frac{\left(1+z\right)^{m+p-1}}{\left(1-z\right)^{p+1}} \end{eqnarray} $$

Here is a verification:

In[27]:= With[{p = 5, 
  m = 7}, {SeriesCoefficient[(1 + z)^(m + 2 p)/(1 - z^2)^(
   p + 1), {z, 0, m}], 
  Sum[Binomial[p + s, s] Binomial[2 p + m, 2 p + 2 s], {s, 
    0, \[Infinity]}]}]

Out[27]= {71808, 71808}

Let's continue: $$ \begin{eqnarray} [z]^m \frac{\left(1+z\right)^{m+p-1}}{\left(1-z\right)^{p+1}} &=& \sum_{s=0}^\infty \binom{p+m-1}{m-s} \binom{p+s}{s} = \sum_{s=0}^\infty \binom{p+m-1}{p+s-1} \binom{p+s}{s}\\ &=& \sum_{s=0}^\infty \frac{(p+s) (m+p-1)!}{p! s! (m-s)!} = \sum_{s=0}^\infty \frac{p (m+p-1)!}{p! s! (m-s)!} + \sum_{s=0}^\infty \frac{s (m+p-1)!}{p! s! (m-s)!} \\ &=& \binom{m+p-1}{m} \left( \sum_{s=0}^\infty \binom{m}{s} + \sum_{s=0}^\infty \frac{s}{p} \binom{m}{s} \right) \\ &=& \binom{m+p-1}{m} \left( 2^m + 2^{m-1} \frac{m}{p} \right) \end{eqnarray} $$

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Leave some scraps for the rest of us ;-) I was about to post a negative binomial solution, but now it is redundant. –  robjohn Nov 1 '11 at 23:29
    
@Sasha, in your last equation you got $(2^m + 2^{m-1} \frac{m}{p}) = 2^{m-1}(\frac{2p+m}{p})$. But i have $2^{m-1}\frac{2p+m}{m}$ in my identity. –  Philipp G. Sinicyn Nov 3 '11 at 18:35
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@PhilippG.Sinicyn Yes, but you also have a different binomial. Notice that $\frac{1}{m} \binom{m+p-1}{p} = \frac{1}{p} \binom{m+p-1}{m}$. –  Sasha Nov 3 '11 at 18:39

Let $d_s = \binom{p+s}{s} \binom{2p+m}{2p+2s}$. Using the recurrence relations for binomial, the ratio of successive terms is: $$ \frac{d_{s+1}}{d_s} = \frac{\left(s - m/2\right)\left(s -(m-1)/2\right)}{ (s+1)(s+p+1/2) } = \frac{(s+a)(s+b)}{(s+1)(s+c)} $$ The hypergeometric certificate above means that $$ \sum_{s=0}^\infty d_s = d_0 \sum_{s=0}^\infty \frac{(a)_s (b)_s}{s! (c)_s} = \binom{2p+m}{2p} {}_2 F_1\left( -\frac{m}{2}, -\frac{m-1}{2} ; p+\frac{1}{2} ; 1\right) $$ where $a = -\frac{m}{2}$, $b=-\frac{m-1}{2}$ and $c=p+\frac{1}{2}$.

Using Gauss's theorem, valid for $c>a+b$: $$ {}_2 F_1\left( a, b; c; 1\right) = \frac{\Gamma(c) \Gamma(c-a-b)}{\Gamma(c-a) \Gamma(c-b)} $$ we obtain the required identity: $$ \sum_{s=0}^\infty \binom{p+s}{s} \binom{2p+m}{2p+2s} = \binom{2p+m}{2p} \frac{\Gamma\left(p+\frac{1}{2}\right) \Gamma\left( p+m \right)}{ \Gamma\left( p+\frac{m+1}{2} \right) \Gamma\left( p+\frac{m}{2} \right) } $$ Applying the duplication formula for $\Gamma(2p+m+1)$ and $\Gamma(2p+1)$ arising from $\binom{2p+m}{2p}$ we arrive at the result: $$ \sum_{s=0}^\infty \binom{p+s}{s} \binom{2p+m}{2p+2s} = 2^{m-1} (m+2p) \frac{\Gamma(m+p)}{\Gamma(m+1) \Gamma(p+1)} = 2^{m-1} \frac{m+2p}{m+p} \binom{m+p}{p} $$

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Thank you, but i think i need to prove it using only generating functions and formal power series. –  Philipp G. Sinicyn Nov 1 '11 at 21:41
    
@PhilippG.Sinicyn I have posted another answer that uses generating functions –  Sasha Nov 1 '11 at 22:46

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