Nice. I figured out sum(1..n) = n(n+1)/2 in a slightly way (well, sum(0..n-1) = n(n-1)/2). My grandfather played chess in a club, and he had these sheets of game results where everybody played everybody else once. I wondered how many blocks there were in the table. Well, it looks like a right triangle of height n and base n (which has area nn/2) but with n little triangles notched out (total area n/2). So the area of the blocks is (nn-n)/2, and each has area 1, so there are n(n-1)/2 of them.