Combinatoric meaning of the sum $1\cdot2+1\cdot3+1\cdot4+\ldots+(n-1)\cdot n$ Does anyone know a beautiful meaning (for example, geometric) but not entirely trivial combinatoric meaning of the sum 
$$
\sum_{1\leqslant a<b\leqslant n}ab=1\cdot2+1\cdot3+1\cdot4+\ldots+(n-1)\cdot n\quad?
$$
Thanks! 
 A: It's easy to see that $$(1+2+...+n)^2=1^2+2^2+...+n^2+2S$$ and so $$S=\frac{1}{2}\left [\left ( \sum_\limits{i=1}^{n}i \right )^2 - \sum_\limits{i=1}^{n}i^2 \right]$$
Now, the idea is to see $a^2$ combinatorically as the answer to the following question: 

"you have a blue and a red coin and $a$ different people. On how many
  ways can you give the coins to the people (giving both to the same
  person is allowed)?".

$\sum_\limits{i=1}^{n}i$ is the number of elements in a (2-dimentional) pyramid with base $n$.
Therefore $2S$ is the answer to the question:

There are $1+2+...+n$ people arranged in a pyramid. On how many ways can you give them a blue and a red coin, so that it's forbidden to give both coins to $2$ people in the same row?


Since we now can't give the same person $2$ coins, dividing by $2$ is just ignoring the two distinct identities of the coins, i.e, $S$ is answer to the same problem but with identical coins.
A: Note that your sum can be written as
$$s_n={1\over2}\left(\left(\sum_{k=1}^n k\right)^2-\sum_{k=1}^n k^2\right)\ .$$
This suggests the following combinatorial view:
Imagine a triangular array of discs with $n$ discs in the bottom row, $n-1$ disc in the next row, and so on, ending with a single disc at the top. In how many ways can you mark two of these discs, not both of them in the same horizontal row?
The result is
$$s_n={1\over24}(3n^4+2n^3-3n^2-2n)\ .$$
