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I have to prove this (a combinatorally proof, counting a set in two different ways):

$$\sum_{i=1}^n\sum_{j=1}^n\mathrm{min}(i,j)=\sum_{k=1}^nk^2 .$$

This is what I have done: take the set $$\{(x_1,x_2,x_3)\in\mathbb{Z}^3:1\leq x_1, x_2 \leq k, x_3=k, k=1,\ldots,n\}.$$ This set consists of $n$ squares with increasing side. I noticed that if you count the set adding the points diagonally square by square, you get the formula with the minimum. I mean if $n=3$, then the diagonals with $1$ point are $5$, the diagonals with $2$ points are $3$ and the diagonal with $3$ points is $1$, so you have $$1+1+1+1+1+2+2+2+3.$$ But I can't formalize that, could you help me please?

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up vote 3 down vote accepted

You could visualize it like this. Start with an $n \times n$ square divided into $n^2$ unit squares. Label the rows and the columns $1$ through $n$. Now imagine that you have a bunch of cubical blocks one unit on a side. On the unit square in row $i$, column $j$ you stack $\min\{i,j\}$ of these blocks. Clearly this requires $\sum_{i=1}^n \sum_{j=1}^n \min\{i,j\}$ blocks.

To get the other side of the identity, count the blocks by layers. There are clearly $n^2$ blocks in layer one, at the bottom. The column of blocks in row $i$, column $j$ reaches up to layer $k$ if and only if $k \le \min\{i,j\}$, which is true if and only if $k \le i$ and $k \le j$, i.e., $i,j \ge k$. The number of cells in the original grid satisfying $i,j \ge k$ is $[n-(k-1)]^2$ so counting by layers gives a total of $$\sum\limits_{k=1}^n [n-(k-1)]^2 = \sum\limits_{k=0}^{n-1} (n-k)^2 = \sum\limits_{k=1}^n k^2$$ blocks.

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You could try a recursion. If $L(n)$ is the left side, what is $L(n+1) - L(n)$?

Your observation makes a nice "proof without words", though.

enter image description here

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I have to prove it counting a set in two different ways, but I can't formalize my previous argument – Alex M Sep 6 '11 at 2:32
@Alex: That's exactly what the picture does. If you count the number of cubes layer by layer, you get the right side. If you count the number of cubes by vertical heights, you get the left side. – Ted Sep 6 '11 at 2:45

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