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Math Help - Obtain an expression

  1. #1
    Senior Member
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    Obtain an expression

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

     <br />
\sum^{2n}_{r=n+1}(2r-1)^2<br />


    Thanks .
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  2. #2
    Member Nacho's Avatar
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    <br />
\begin{gathered}<br />
  \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 \\<br />
  \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 \\ <br />
\end{gathered} <br />

    Can you continue?
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  3. #3
    Member Abu-Khalil's Avatar
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    You may also rewrite it as \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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