// So glad it was a simple solution ;-D //
:P
I normally hate using "judgmental words" in my posts -- "simple", "obvious", "clearly", "trivial", etc. -- because they feel quite presumptuous and often the very fact that the question is asked means that the answer *isn't* simple, clear, obvious or whatever. Here it seems appropriate because I'm comparing the answer to what it *could* have been, some horribly large and arbitrary number that you'd have to get by exhaustive counting, or derive in some hideously complex way. So it's comparatively "simple" in that respect: you can deduce the pattern either by thinking about smaller chessboards, as I did it, or by considering a single row and the effect of adding further rows above/below, a la Gizmonster.
A couple of useful extensions:
1. The number of arbitrary rectangles on an n*m chessboard is n(n+1)m(m+1)/4, or just T_n*T_m for triangle numbers T_i = 1,3,6,10...
2. The number of arbitrary squares on an n*n chessboard is equal to the sum of the first n square numbers, and has the formula n(n+1)(2n+1)/6 -- for a standard 8*8 chessboard this comes to 204.
3. The number of arbitrary squares on an n*m chessboard is a formula I'm not aware of having come across before, but turns out to be also relatively compact. Assuming n is smaller than m, then the result is:
No. of squares = n(n+1)(3m-n+1)/6
With m and n reversed in this formula if n is larger. For example, there are 276 squares of various size on a Capablanca chessboard:
https://en.wikipedia.org/wiki/Capablanca_chess