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: solution (spoiler) by Daniel)

If in Windows, try ALT+228 to get Σ (sigma). (This is NOT the same as ALT+0228 which gives ä) -- hope this works.

Posted by ed bottemiller on 2008-09-26 19:17:37