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March 27, 2007

Sum of the squares of the first n integers

Filed under: Mathematics — mrc @ 9:49 am
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So I feel pretty comfortable with the proof by induction of the formula for the sum of the squares of the first n integers:

\displaystyle\sum_{i=1}^{n}i^2 = \frac{n(n+1)(2n+1)}{6}

It’s one thing to prove it’s true, and another thing to have some intuition as to its origin. Who thought this up in the first place? I guess I’m looking for a trick like Gauss has for adding up the sum of the first n integers. When I think about his arrangement of the numbers, the \displaystyle\sum_{i=1}^{n}i = \frac{n(n+1)}{2} formula makes sense. Does anyone know something like that for the sum of the squares? Does any particular person get credit for coming up with the formula?

10 Comments

  1. Check this out:

    http://mathforum.org/library/drmath/view/56920.html

    Comment by e — March 27, 2007 @ 10:05 am

  2. Oh, interesting. It looks like that same approach is discussed here:
    http://www.cut-the-knot.org/htdocs/dcforum/DCForumID3/183.shtml

    I’m not sure what to think about the idea that induction is actually being used here in disguise. I suspect that it is not, because (n+1)2 = n2 + 2n + 1 and therefore (n+1)2 - n2 = 2n + 1 can be shown simply using algebra. Telescoping sums and induction are not the same thing.

    Thanks for the tip!

    Comment by mrc — March 27, 2007 @ 10:20 am

  3. Just last week I used geometry for Gauss (line up little unit squares in a right triangle, base = n, top row = 1, supreimpose a real right triangle, area is (n^2)/2, little extra “teeth” have area 1/2 each, for a total of n/2. Add the pieces and we have (n^2 + n)/2.

    Perhaps there is an extension to 3 dimensions? Note that sum{1,n}(i^2) = (2n+1)/3 x sum{1,n}(i). That 1/3 factor makes me think, pyramid?

    It would be interesting to develop something, or to find it already done.

    Comment by Jonathan — March 27, 2007 @ 6:06 pm

  4. I think that the step pyramid is probably the best visual aid, as the last comment mentioned, and as illustrated here:

    http://mathforum.org/library/drmath/view/53646.html

    Only in reality, the formula for volume of a pyramid/cone is a result of the sum of squares, rather than the other way around.

    Comment by tony lucchese — March 28, 2007 @ 5:37 pm

  5. This is probably more what you’re looking for. It’s attributed to Archimedes, so you know it’s got to be good.

    http://www.macalester.edu/~bressoud/pub/CBN2.pdf

    Comment by tony lucchese — March 29, 2007 @ 5:21 am

  6. Tony, thank you! This is a great resource.

    Comment by mrc — March 29, 2007 @ 9:49 am

  7. Here’s an animation showing a geometric approach:
    http://www.artofproblemsolving.com/Images/Flash/Side7.swf

    Comment by mrc — March 31, 2007 @ 11:15 pm

  8. Can any one proove sum of the squares of the first n integers is 1/24 (2n) (2n+1) (2n+2) step by step instructions pls

    Comment by Sunny — April 1, 2007 @ 4:58 am

  9. Sunny: You can use induction to prove your formula. I suggest you do a bit more reading in your text or research online if you’re not sure what that means.

    Comment by mrc — April 1, 2007 @ 1:30 pm

  10. There’s a simple, intuitive approach to discovering the formula in Richard Friedman’s quirky but delightful book, _An Adventurer’s Guide to Number Theory_, pp. 32-34. It relies on seeing patterns when you factor the first few sums–1, 5, 14, 30, 55, 91, etc. It’s really nice! (The actual proof of the formula is by induction and comes later.)

    By the way, I found it an interesting little exercise to prove that [n(n+1)(2n+1)]/6 is always an integer for every n.

    Comment by Paul Frommer — August 18, 2007 @ 5:41 pm

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