Simplify a triple sum I need to find a closed form for this summation:
$$\sum_{j=1}^m\sum_{i=j}^m\sum_{k=j}^m\frac{{m\choose i}{{m}\choose{k}}}{j{m\choose j}}r^{k-j+i}$$
I posted this a long time ago, but today I found out there was an important typo in the formula, so I repost the correct one again here.
Any minor simplification or closed form for the special case of $r=1$ is also helpful.
Any help is greatly appreciated!
 A: @Azoodish;
I had Mathematica run it off since I could not do anything with it even with r = 1.
For r = 1.
$-2^{m-1} \left(H_m+2^{m+1} \Phi (2,1,m+1)+i \pi \right)$
$H_m$
is the mth Harmonic number. The other one is the LerchPhi function which is defined here:
http://mathworld.wolfram.com/LerchTranscendent.html
I know, not that useful, but at least you know there is some answer or maybe that a simple closed form does not exist. I did some numerical testing of this solution, up to m = 1000. The usual warnings go with a result like this and it is up to you to prove the result.
A: Take a good look at the inner sum. The only k-dependent term is ${m\choose k}r^k$, and we are basically asked to find a closed form for what is obviously a partial expansion of $(r+1)^m$. But it is known, and has already been asked before on this site quite a few times, that such a form does not exist, not even for the simple case $r=1$. In other words, there is no simple expression for the partial sum of binomial coefficients. I'm not saying that you cannot re-write this sum in terms of, say, Gaussian hypergeometric functions $_2F_1$ and the like, but these are not regarded as closed form expressions.
