$\sum_{j=2}^{n}\sum_{k=1}^{j-1}kj=\frac{1}{24}(x-1)x(x+1)(3x+2)$

Question

How can I get from $\sum_{j=2}^{n}(\sum_{k=1}^{j-1}kj)$ the equation $\frac{1}{24}(x-1)x(x+1)(3x+2)$

Who would the equation be the following case?

$\sum_{j=3}^{n}(\sum_{m=2}^{j-1}(\sum_{k=1}^{m-1}kmj))=?$

Answer 1

By the Hockey-Stick identity: $$\sum_{j=2}^{n}j\sum_{k=1}^{j-1}k = \sum_{j=2}^{n}j\binom{j}{2}=3\sum_{j=2}^{n}\binom{j}{3}+2\sum_{j=2}^{n}\binom{j}{2}=3\binom{n+1}{4}+2\binom{n+1}{3} $$ and the RHS of the last identity can also be written as $\frac{3n+2}{4}\binom{n+1}{3}$.

In a similar way, $$ \begin{eqnarray*}\sum_{j=3}^{n}\sum_{m=2}^{j-1}\sum_{k=1}^{m-1}kmj = \sum_{j=3}^{n}\sum_{m=2}^{j-1}jm\binom{m}{2}&=&3\sum_{j=3}^{n}\sum_{m=2}^{j-1}j\binom{m}{3}+2\sum_{j=3}^n\sum_{m=2}^{j-1}j\binom{m}{2}\\&=&3\sum_{j=3}^{n}j\binom{j}{4}+2\sum_{j=3}^{n}j\binom{j}{3}\\&=&\sum_{j=3}^{n}\left[15\binom{j}{5}+12\binom{j}{4}+8\binom{j}{4}+6\binom{j}{3}\right]\\&=&15\binom{n+1}{6}+20\binom{n+1}{5}+6\binom{n+1}{4}.\end{eqnarray*}$$

As suggested in the comments, shortcuts are given by Stirling numbers of the first kind, since we are dealing with the elementary symmetric polynomials of $\{1,2,\ldots,n\}$.