• Show that the sum of the first n squares is n(n+1)(2n+1)/6.
  • I know this is often in the textbook for proof by induction, which is why proof by induction is not allowed.

This is a relatively hard one, take your time.

  • siriusmart@lemmy.worldOPM
    3·
    6 months ago

    i took the second setup cuz thats what i saw when thinking of the problem, i’ll read the first approach later

    • zkfcfbzr@lemmy.worldEnglish
      3·
      6 months ago

      I did see a third setup where you take Σ (i = 1 to n) (Σ (k = 1 to i) 2k-1), which is a valid setup where you count all the blocks on the top of their column, then remove those and count the new top blocks, etc - but it ended up not leading to a proof of the formula because as it turns out, there are always i^2 (from i = n to 1) blocks on the top layer. So it’s just a worse way of doing Σi^2 and we get a 0 = 0 situation.