# what's the summation of this finite sequence?

$a$ 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.

• What do you mean by "physical meaning"? Commented Nov 16, 2014 at 2:26
• Some intuition behind the solution. How to understand it intuitively.:-) Commented Nov 16, 2014 at 2:27

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 $k = b+1$.

• It makes sense! You are so smart! Could you give some physical ideas behind this equation? Many THANKS! Commented Nov 16, 2014 at 2:24
• 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. Commented Nov 16, 2014 at 2:31
• It makes sense. Do you know the name of the formula? Commented Nov 16, 2014 at 2:33
• 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. Commented Nov 16, 2014 at 2:35
• I'm not sure it has a name. It is identity #8 on the Wikipedia page for binomial coefficients. Commented Nov 16, 2014 at 2:36

Some experimentation gives $\dbinom{a+b+1}{b+2}$. This is the correct answer for $1\le a, b\le 5$.

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: $$\frac{\Gamma (a+b+2)}{\Gamma (a) \Gamma (b+3)}$$

• Cool! Could you give some analytical ideas? Commented Nov 16, 2014 at 2:15
• @Fred: Mine didn't include the $x$... * embarrassed noises *. Thanks :) Commented Nov 16, 2014 at 2:32