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Math Help - Sigma notation

  1. #1
    Member SengNee's Avatar
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    Sigma notation

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


    No need show me the method, just give me the answer, thanks.
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  2. #2
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    Quote Originally Posted by SengNee View Post
    \sum_{r=1}^{n}\frac {2r}{(2r-1)(2r+1)(2r+3)}


    No need show me the method, just give me the answer, thanks.
    From the usual partial fraction decomposition:


    \frac {2r}{(2r-1)(2r+1)(2r+3)} = \frac{1}{8} \left( \frac{1}{2r - 1}\right) + \frac{1}{4} \left( \frac{1}{2r + 1}\right) - \frac{3}{8} \left( \frac{1}{2r + 3}\right).


    So the for n > 2, the first few terms of the series are:


    \frac{1}{8} \left( 1 + \frac{1}{3} + \frac{1}{5} + ....\right) + \frac{1}{4} \left( \frac{1}{3} + \frac{1}{5} + .....\right) - \frac{3}{8} \left( \frac{1}{5} + \frac{1}{7} + ....\right).


    Note that \frac{1}{8} + \frac{1}{4} - \frac{3}{8} = 0 ......


    So what terms will be left if n = 3? n = 4? n = n?

    If I was less lazy I'd see this through to the answer, which I would then give you .....
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by SengNee View Post
    No need show me the method, just give me the answer, thanks.
    As this is a Math *HELP* Forum, I find this comment to be amazing. Why would you just want the answer and not learn how to do it?

    -Dan
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  4. #4
    Member SengNee's Avatar
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    Quote Originally Posted by mr fantastic View Post
    From the usual partial fraction decomposition:


    \frac {2r}{(2r-1)(2r+1)(2r+3)} = \frac{1}{8} \left( \frac{1}{2r - 1}\right) + \frac{1}{4} \left( \frac{1}{2r + 1}\right) - \frac{3}{8} \left( \frac{1}{2r + 3}\right).


    So the for n > 2, the first few terms of the series are:


    \frac{1}{8} \left( 1 + \frac{1}{3} + \frac{1}{5} + ....\right) + \frac{1}{4} \left( \frac{1}{3} + \frac{1}{5} + .....\right) - \frac{3}{8} \left( \frac{1}{5} + \frac{1}{7} + ....\right).


    Note that \frac{1}{8} + \frac{1}{4} - \frac{3}{8} = 0 ......


    So what terms will be left if n = 3? n = 4? n = n?

    If I was less lazy I'd see this through to the answer, which I would then give you .....
    n=n
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  5. #5
    Member SengNee's Avatar
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    Quote Originally Posted by topsquark View Post
    As this is a Math *HELP* Forum, I find this comment to be amazing. Why would you just want the answer and not learn how to do it?

    -Dan
    Because I had know how to do and I did it step by step, but my answer is different to the book's answer.
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by SengNee View Post
    Because I had know how to do and I did it step by step, but my answer is different to the book's answer.
    Then it's probably best to at least post your solution to see who agrees with it.

    -Dan
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