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Series Question

  1. Sep 29, 2005 #1
    I've got a very tricky series question on hand,

    I'm given [tex] \sum_{r=1}^{\infty} r^2 = \frac{1}{6} n(n+1)(2n+1) [/tex] and i'm asked to find [tex] \sum_{r=n+1}^{2n} 4r^2 [/tex].

    The change of limits and all is killing me ! any ideas on how to start? Thanks.
  2. jcsd
  3. Sep 29, 2005 #2


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    Science Advisor

    There's an obvious typo in the "given" formula- there is no "n" on the left hand side. It should be
    [tex] \sum_{r=1}^n r^2 = \frac{1}{6} n(n+1)(2n+1) [/tex]

    First, obvious point: ignore the "4" until you have done just [tex] \sum_{r=n+1}^{2n} r^2 [/tex], then multiply by 4.

    Can you use the "given" formula to find [tex] \sum_{r=1}^{2n} r^2 [/tex]?
    Can you use the "given" formula to find [tex] \sum_{r=1}^{n} r^2 [/tex]?

    What is the difference between them?
    Last edited by a moderator: Sep 29, 2005
  4. Sep 29, 2005 #3
    OH yes ! thanks alot for the help, and esp. pointing out that typo. Am able to solve now.

    : )
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