The following appears to be true. Let $n \geq 1$ be an integer. Then

$$ \sum_{i=1}^n \frac{n! (n-1)! }{(n-i)! (n+i+1)!}i (i+1) (2i+1) = 1.$$

This cropped up when doing some calculations in [coalescent theory](http://en.wikipedia.org/wiki/Coalescent_theory). I found the distribution of the time of a particular event as a mixture of exponentials. I can calculate the weights numerically and the weights appear to be the terms in the above sum.

I also noticed:

* The last term ($i=n$) is the reciprocal of a Catalan number.

* The sum of alternate terms ($i$ odd or $i$ even) appears to be always 1/2.

* $i (i+1) (2i+1) / 6$ is the sum of the first $i$ squares.

Something is going on that I don't understand! Can anyone shed any light?

$$ \sum_{i=1}^n \frac{n! (n-1)! }{(n-i)! (n+i+1)!}i (i+1) (2i+1) = 1.$$

This cropped up when doing some calculations in [coalescent theory](http://en.wikipedia.org/wiki/Coalescent_theory). I found the distribution of the time of a particular event as a mixture of exponentials. I can calculate the weights numerically and the weights appear to be the terms in the above sum.

I also noticed:

* The last term ($i=n$) is the reciprocal of a Catalan number.

* The sum of alternate terms ($i$ odd or $i$ even) appears to be always 1/2.

* $i (i+1) (2i+1) / 6$ is the sum of the first $i$ squares.

Something is going on that I don't understand! Can anyone shed any light?