sum of a positive series

Question

For sufficiently large $m$, I want to calculate the following series.

$\displaystyle\sum_{i=0}^{3m}\sum_{j=0}^m \sum_{k=0}^m\sum_{l=0}^m \min(i,j+k+l)$

Except considering all case by case, is there any clever trick?

Answer 1

Hint: \begin{eqnarray*} \sum_{i=0}^{3m}\sum_{j=0}^m \sum_{k=0}^m\sum_{l=0}^m \min(i,j+k+l) &=& \sum_{j,k,l=0}^m \left[\sum_{i=0}^{j+k+l} i + \sum_{i=j+k+l+1}^{3m}(j+k+l) \right]\\ &=& \sum_{j,k,l=0}^m \bigg[ \frac{1}{2}(j+k+l)(j+k+l+1) \\ &&+ (j+k+l)(3m-j-k-l)\bigg] \end{eqnarray*}

The sum can now be found using only \begin{eqnarray*} \sum_{k=0}^n 1 &=& n \\ \sum_{k=0}^n k &=& \frac{n(n+1)}{2} \\ \sum_{k=0}^n k^2 &=& \frac{n(n+1)(2n+1)}{6}. \end{eqnarray*}