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Thread: Obtain an expression

  1. #1
    Senior Member
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    Obtain an expression

    Given that $\displaystyle \sum^{n}_{r=1}r^2=\frac{1}{6}n(n+1)(2n+1)$ , obtain an expression , in its simplest term , for

    $\displaystyle
    \sum^{2n}_{r=n+1}(2r-1)^2
    $


    Thanks .
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  2. #2
    Member Nacho's Avatar
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    $\displaystyle
    \begin{gathered}
    \sum\limits_{r = n + 1}^{2n} {\left( {2r - 1} \right)^2 } = \sum\limits_{r = 1}^n {\left( {2(r + n) - 1} \right)^2 } = \sum\limits_{r = 1}^n {\left( {2r + 2n - 1} \right)^2 } = \hfill \\
    \sum\limits_{r = 1}^n {4r^2 + 4n^2 + 1 + 8rn - 4r - 4n} = 4\sum\limits_{r = 1}^n {r^2 } + \left( {4n^2 - 4n + 1} \right)\sum\limits_{r = 1}^n 1 + \left( {8n - 4} \right)\sum\limits_{r = 1}^n r \hfill \\
    \end{gathered}
    $

    Can you continue?
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  3. #3
    Member Abu-Khalil's Avatar
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    You may also rewrite it as $\displaystyle \sum_{k=n+1}^{2n}(2r-1)^2=\sum_{k=1}^{2n}(2r-1)^2-\sum_{k=1}^{n}(2r-1)^2$, expand the cuadratic binomy and apply your formula.
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