1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    HallsofIvy

    User Avatar
    Staff Emeritus
    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: 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.

    : )
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Series Question
  1. A series question (Replies: 4)

  2. Series of questions (Replies: 4)

Loading...