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$a$ and $b$ are positive integers. The summation is $$\sum\limits_{x = 1}^a {x\left( {\begin{array}{*{20}{c}} {a + b - x}\\ b \end{array}} \right)} .$$ Any closed-form expression?

I thought it should have. And maybe there is some physical meaning behind it.


Sorry, I've simplified the problem, and now it becomes easier.

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  • $\begingroup$ What do you mean by "physical meaning"? $\endgroup$ – Eric Stucky Nov 16 '14 at 2:26
  • $\begingroup$ Some intuition behind the solution. How to understand it intuitively.:-) $\endgroup$ – Stupid_Guy Nov 16 '14 at 2:27
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This is the binomial identity $\sum_{m=0}^n\binom{m}{j}\binom{n-m}{k-j} = \binom{n+1}{k+1}$ with $j = 1$, $n = a+ b$ and $k = b+1$.

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  • $\begingroup$ It makes sense! You are so smart! Could you give some physical ideas behind this equation? Many THANKS! $\endgroup$ – Stupid_Guy Nov 16 '14 at 2:24
  • $\begingroup$ Think of the ways of choosing $k+1$ objects out of $n+1$. In one way of counting it is simply $\binom{n+1}{k+1}$, and in another you fix $j$, and count the ways to choose the objects so that the $m+1$st object is the $j+1$st chosen. $\endgroup$ – Michael Biro Nov 16 '14 at 2:31
  • $\begingroup$ It makes sense. Do you know the name of the formula? $\endgroup$ – Stupid_Guy Nov 16 '14 at 2:33
  • $\begingroup$ I've decided to give you the bounty. But the system notifies me that I cannot operate until 24 hours later. You are the first, I will operate it tomorrow. $\endgroup$ – Stupid_Guy Nov 16 '14 at 2:35
  • $\begingroup$ I'm not sure it has a name. It is identity #8 on the Wikipedia page for binomial coefficients. $\endgroup$ – Michael Biro Nov 16 '14 at 2:36
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Some experimentation gives $\dbinom{a+b+1}{b+2}$. This is the correct answer for $1\le a, b\le 5$.

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Wolfram Alpha produces:$$\sum_{x=1}^a x \binom{a+b-x}{b} =\frac{(a+b) (a+b+1) \binom{a+b-1}{b}}{(b+1) (b+2)}$$ Full simplification of RHS: $$\frac{\Gamma (a+b+2)}{\Gamma (a) \Gamma (b+3)}$$

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  • $\begingroup$ Cool! Could you give some analytical ideas? $\endgroup$ – Stupid_Guy Nov 16 '14 at 2:15
  • $\begingroup$ @Fred: Mine didn't include the $x$... * embarrassed noises *. Thanks :) $\endgroup$ – Eric Stucky Nov 16 '14 at 2:32

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