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Summation of i Squared

  1. Sep 4, 2011 #1
    1. The problem statement, all variables and given/known data
    I'm not very proficient with LAtex, so I'll try to translate this mess the best I can. It's a summation
    [itex]\sum[/itex] i2 (on the bottom, there would be an "i = 1") (on the top, there would be an "n") this summation equals [itex]\frac{n(n+1)(2n+1)}{6}[/itex]

    In the summation, basically, "i" starts off at one and goes to "n"

    Why does this summation equal [itex]\frac{n(n+1)(2n+1)}{6}[/itex]?

    2. Relevant equations
    The definition of a definite integral? I found this summation through doing definite integrals using Riemann sums. I found the answer to the summation online, but I wanted to know how one arrives at the answer.

    3. The attempt at a solution
    So far, I have done this:

    S = 12 + 22 + 32 + ... + (n - 1)2 + n2
    S = n2 + (n - 1)2 + (n - 2)2 + ... + 22 + 12

    I just wrote out a part of the summation, and under it, I did the same summation part but in reverse.

    Then...I added the two summations to get

    2S = n2 + 1 + (n - 1)2 + 4 + (n - 2)2 + 9 + ... + (n - 1)2 + 4 + n2 + 1

    I saw somewhat of a pattern, but it was hard to explain and I had to stop here. If someone could give me a link to an explanation, or if someone could fit an explanation in a response or two as to why this summation equals [itex]\frac{n(n+1)(2n+1)}{6}[/itex], that would be fantastic. Thank you in advance!
     
  2. jcsd
  3. Sep 4, 2011 #2

    eumyang

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