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Math Help - Confusion in Series

  1. #1
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    Confusion in Series

    Given that Ur = r (2r + 1) + 2^{r + 2} , find
    n
    Ur
    r=1

    I’ve done but I’m confused about my answer. Please tell me is it wrong or right.
    I got  \frac{1}{6} n(n + 1) (4n + 5) + 2^{(n + 3)} - 4
    Last edited by geton; January 23rd 2008 at 09:37 PM. Reason: Correcting 2r + 2
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  2. #2
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    Quote Originally Posted by geton View Post
    Given that Ur = r (2r + 1) + 2r + 2 , find
    n
    Ur
    r=1

    I’ve done but I’m confused about my answer. Please tell me is it wrong or right.
    I got  \frac{1}{6} n(n + 1) (4n + 5) + 2^{(n + 3)} - 4
    U_r = r(2r + 1) + 2r + 2 = 2r^2 + 3r + 2.

    Therefore \sum_{r=1}^{n} U_r = \sum_{r=1}^{n} (2r^2 + 3r + 2) = 2 \sum_{r=1}^{n}r^2 + 3 \sum_{r=1}^{n} r + 2n

    Now apply the standard formulae for the sum of first n integers and sum of first n squares of integers ......
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  3. #3
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    Iím really very sorry for my mistake.
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  4. #4
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    \frac{1}{6} n(n + 1) (4n + 5) + 2^{(n + 3)} - 4
    Looks right to me.
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