[eru]

This thread is pedantic.

By thy way:

  sum(i=1..n) i*i = n/6 + n^2/2 + n^3/3

[jgrahamc]

Proof?

[hhm]

This proof helps: http://pirate.shu.edu/~wachsmut/ira/infinity/answers/sm_sq_c...

From that: sum bla= n(n+1)(2n+1)/6, then you go distribute

[eru]

Nice. How did I get the result in the first place?

I know that this sum can be expressed as a cubic polynomial. So I just evaluated the sum at four data-points (n=0,1,2,3) and used those to determine the four coefficents by solving a linear equation system.

[jgrahamc]

Turns out that drawing a picture really helps: http://news.ycombinator.com/item?id=101532

[eru]

You can usually come up with a clever picture only after you understood the problem - not before.