[Math] How to calculate this triple summation

Question

I need to calculate the following summation:
$$\sum_{j=1}^m\sum_{i=j}^m\sum_{k=j}^m\frac{{m\choose i}{{m-j}\choose{k-j}}}{k\choose j}r^{k-j+i}$$
I do not know if it is a well-known summation or not.

(The special case when $r=1$ is also helpful.)

Even a little simplification is good, unfortunately I cannot simplify it more than this!

Edit: another way to write this summation is:
$$\sum_{j=1}^m\sum_{i=j}^m\sum_{k=j}^m\frac{{m\choose i}{{m}\choose{k}}}{j{k\choose j}}r^{k-j+i}$$
Anybody can help with this one?

Answer 1

Hint: This is just a starter which might be helpful for further calculation. We obtain a somewhat simpler representation of the triple sum.

\begin{align*} \sum_{j=1}^m&\sum_{i=j}^m\sum_{k=j}^m\frac{\binom{m}{i}\binom{m-j}{k-j}}{\binom{k}{j}}r^{k-j+i}\\ &=\sum_{j=1}^m\sum_{i=j}^m\sum_{k=j}^{m}\frac{m!}{i!(m-i)!}\cdot\frac{(m-j)!}{(k-j)!(m-k)!}\cdot\frac{j!(k-j)!}{k!}r^{k-j+i}\tag{1}\\ &=\sum_{j=1}^m\sum_{i=j}^m\sum_{k=j}^{m}\frac{\binom{m}{k}\binom{m}{i}}{\binom{m}{j}}r^{k-j+i}\tag{2}\\ &=\sum_{j=1}^m\frac{1}{\binom{m}{j}r^j}\sum_{i=j}^m\binom{m}{i}r^i\sum_{k=j}^m\binom{m}{k}r^k\tag{3}\\ &=\sum_{j=1}^m\frac{1}{\binom{m}{j}r^j}\left(\sum_{k=j}^m\binom{m}{k}r^k\right)^2 \end{align*}

Comment:

• In (1) we write the binomial coefficients using factorials

• In (2) we cancel out $(k-j)!$ and rearrange the other factorials to binomial coefficients so that each of them depends on one running index only

• In (3) we can place the binomial coefficients conveniently and see that the sums with index $i$ and $k$ are the same