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Math Help - series problem .

  1. #1
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    series problem .

    Find the sum to the n terms of the series :

     \frac{1}{(2r-1)(2r+1)(2r+3)}
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  2. #2
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    Sum of series

    Hello thereddevils
    Quote Originally Posted by thereddevils View Post
    Find the sum to the n terms of the series :

     \frac{1}{(2r-1)(2r+1)(2r+3)}
    Using Partial Fractions, you find that

    \frac{1}{(2r-1)(2r+1)(2r+3)} = \frac{1}{8}\left(\frac{1}{2r-1}-\frac{2}{2r+1}+\frac{1}{2r+3}\right)

    So if the sum to n terms is S:

    8S = \sum_{r=1}^n\left(\frac{1}{2r-1}-\frac{2}{2r+1}+\frac{1}{2r+3}\right)

    = \sum_{r=1}^n\frac{1}{2r-1} -\sum_{r=2}^{n+1}\frac{2}{2r-1}+\sum_{r=3}^{n+2}\frac{1}{2r-1}

     = 1 + \frac{1}{3}+\sum_{r=3}^n\frac{1}{2r-1}-(\frac{2}{3}+2\sum_{r=3}^n\frac{1}{2r-1} + \frac{2}{2n+1})+\sum_{r=3}^n\frac{1}{2r-1}+\frac{1}{2n+1}+\frac{1}{2n+3}

    =\frac{2}{3}-\frac{2}{2n+1}+\frac{1}{2n+1}+\frac{1}{2n+3}

    \Rightarrow S=\frac{1}{8}\left(\frac{2}{3}-\frac{1}{2n+1}+\frac{1}{2n+3}\right)

    Grandad
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