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Math Help - Sum of the series

  1. #1
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    Sum of the series

    ∑_(r=1)^n▒2/r(r+1)(r+2)
    ∑_(r=1)^n▒r/((2r-1)(2r+1)(2r+3))

    I first convert both to partial fraction.
    Then i use the method of differences to cancel off term by term, but i don't what term should i cancel.

    Guide me through these two questions, please. Thank you.
    Last edited by alexander9408; September 7th 2012 at 12:28 PM.
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  2. #2
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    Re: Sum of the series

    Quote Originally Posted by alexander9408 View Post
    ∑_(r=1)^n▒2/r(r+1)(r+2)
    ∑_(r=1)^n▒r/((2r-1)(2r+1)(2r+3))
    I first convert both to partial fraction.
    Then i use the method of differences to cancel off term by term, but i don't what term should i cancel.
    \sum\limits_{r = 1}^n {\frac{2}{{r\left( {r + 1} \right)\left( {r + 2} \right)}}}  = \sum\limits_{r = 1}^n {\left[ {\frac{1}{r} - \frac{2}{{r + 1}} + \frac{1}{{r + 2}}} \right]}

    Now take n=5 write out all five terms and look for a pattern.
    You may have to do that for several values.
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  3. #3
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    Re: Sum of the series

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

    n=2\,\, \Rightarrow \,\, [(1-\frac{1}{2})+(\frac{1}{3}-\frac{1}{2})]+[(\frac{1}{2}-\frac{1}{3})+(\frac{1}{4}-\frac{1}{3})] =\frac{1}{2}-\frac{1}{3}+\frac{1}{4}

    n=3\,\, \Rightarrow \,\, [(1-\frac{1}{2})+(\frac{1}{3}-\frac{1}{2})]+[(\frac{1}{2}-\frac{1}{3})+(\frac{1}{4}-\frac{1}{3})]+[(\frac{1}{3}-\frac{1}{4})+(\frac{1}{5}-\frac{1}{4})]=\frac{1}{2}-\frac{1}{4}+\frac{1}{5}

    n=4\,\, \Rightarrow \,\, [(1-\frac{1}{2})+(\frac{1}{3}-\frac{1}{2})]+[(\frac{1}{2}-\frac{1}{3})+(\frac{1}{4}-\frac{1}{3})]+[(\frac{1}{3}-\frac{1}{4})+(\frac{1}{5}-\frac{1}{4})]+ [(\frac{1}{4}-\frac{1}{5})+(\frac{1}{6}-\frac{1}{5})]=\frac{1}{2}-\frac{1}{5}+\frac{1}{6}

    \sum_{r=1}^{n}[(\frac{1}{r}-\frac{1}{r+1})+(\frac{1}{r+2}-\frac{1}{r+1})]=\frac{1}{2}-\frac{1}{n+1}+\frac{1}{n+2}
    Thanks from alexander9408
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