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Geometric Sum (Power Series) Calculation

  1. Mar 28, 2008 #1
    Hi

    In trying to calculate the following sum:

    [tex]
    \sum_{i=1}^n{i^2}
    [/tex]

    I found the following expansions:

    [tex]
    \sum_{i=1}^p \sum_{j=0}^{i-1} (-1)^j(i-j)^p {n+p-i+1\choose n-i} {p+1\choose j}
    [/tex]

    My question is: is there an easier or more intuitive way to compute the limit of the sum above?
     
  2. jcsd
  3. Mar 28, 2008 #2
    Yes, it is infinity. :smile:

    And what you have is not a geometric series.
     
    Last edited: Mar 28, 2008
  4. Mar 28, 2008 #3

    tiny-tim

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    Homework Helper

    Hi rwinston! :smile:

    Hint: what is [tex]\sum_{i=1}^n{\left((i+1)^3\,-\,i^3}\right)[/tex] ? :smile:
     
  5. Mar 28, 2008 #4
  6. Mar 28, 2008 #5
    Hi tim

    So I think the expansion of that sum is

    [tex]
    \sum_{i=1}^n{i^3+3i^2+3i+1} - \sum_{i=1}^n{i^3}
    [/tex]
    [tex]
    =\sum_{i=1}^{n}{3i^2+3i+1}
    [/tex]
    [tex]
    =3\sum_{i=1}^n{i^2}+3\sum_{i=1}^n{i}+n
    [/tex]

    Im not sure what the connection is yet tho...
     
    Last edited: Mar 28, 2008
  7. Mar 28, 2008 #6

    tiny-tim

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    ah… but you know what [tex]3\sum_{i=1}^n{i}\,+\,n[/tex] is;

    And you also know what [tex]\sum_{i=1}^n{\left((i+1)^3\,-\,i^3}\right)[/tex] is … don't you … ?

    So [tex]\sum_{i=1}^n{i^2}[/tex] is one-third the difference! :smile:
     
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