recurrence relation expanding $ij$ I need to solve this:
$\displaystyle\sum\limits_{i=1}^n$$\displaystyle\sum\limits_{j=1}^n $$\displaystyle\sum\limits_{k=1}^{i\cdot j} 1$
How do I expand the $i\cdot j$ part? Am I right to do it this way?


*

*$\displaystyle\sum\limits_{i=1}^n$$\displaystyle\sum\limits_{j=1}^n (i\cdot j)$

*$\displaystyle\sum\limits_{i=1}^n i\cdot n (n(n+1)/2)$
And how do I continue from here? The summation quickly gets quite messy.
Thanks!
 A: Whenever you see something not in your summation, you can pull it to the front.
For instance, in your second step, $\displaystyle\sum_{i=1}^n\big[\sum_{j=1}^n(i\cdot j)\big]$, $i$ has nothing to do with summing $j$, so regard it as a constant and factor it out. Therefore, it equals
$$\sum_{i=1}^n\big[i\sum_{j=1}^nj\big]=\sum_{i=1}^ni\frac{n(n+1)}{2}=\frac{n(n+1)}{2}\sum_{i=1}^ni=\left(\frac{n(n+1)}{2}\right)^2$$
A: \begin{eqnarray*}\displaystyle\sum\limits_{i=1}^n\displaystyle\sum\limits_{j=1}^n \displaystyle\sum\limits_{k=1}^{i\cdot j} 1&=&\sum_{i=1}^n \sum_{j=1}^n ij\\
&=& \sum_{i=1}^n  i\left( \sum_{j=1}^n j\right)\\
&=&\left(\sum_{i=1}^n i\right)\left(\sum_{j=1}^n j\right)\\
&=&(\frac{n(n+1)}{2})^2\\
&=&\frac{n^2(n+1)^2}{4}.
\end{eqnarray*}
A: $$\sum_{i=1}^n\sum_{j=1}^n\sum_{k=1}^{ij} 1=\sum_{j=1}^n\sum_{k=1}^{j} 1+\sum_{j=1}^n\sum_{k=1}^{2j} 1+...+\sum_{j=1}^n\sum_{k=1}^{nj} 1=$$
$$\sum_{j=1}^n j+\sum_{j=1}^n 2j+...+\sum_{j=1}^n nj=(1+2+...+n)\sum_{j=1}^n j=$$
$$=(1+2+...+n)(1+2+...+n)=(1+2+...+n)^2$$
