The sum of n+1 consecutive squares, beginning with the square of
n(2n+1), is equal to the sum of the squares of the next n consecutive integers.

   n=1      32 + 42 = 52
   n=2      102 + 112 + 122 = 132 + 142
      .
      .
      .

Prove that the proposition holds for all integers greater than zero.

(In reply to re(2): solution (spoiler) -- print SIGMA by ed bottemiller)

thanks for the tip, I'll remember that in the future.  I had tried using the special characters button at the top to place the sigma symbol.  I'll use ALT+228 from now on. 

Posted by Daniel on 2008-09-26 19:27:47