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Math Help - method of differences

  1. #1
    Senior Member
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    method of differences

    Find its sum :

    \sum^{n}_{r=1}\frac{1}{(3n-2)(3n+1)(3n+4)}

    After doing the partial fraction :

    =\frac{1}{18}\sum^{n}_{r=1}(\frac{1}{3n-2}-\frac{1}{3n+1}+\frac{1}{3n+4})

    Then from here , i tried putting f(r)=\frac{1}{3n-2} , and f(r+1)=\frac{1}{3n+1} .

    Putting these functions back into the summation doesn't work because there is an extra term there .

    What can i do from here ?
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  2. #2
    MHF Contributor red_dog's Avatar
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    You better split the fraction this way:

    \frac{1}{(3n-2)(3n+1)(3n+4)}=\frac{A}{(3n-2)(3n+1)}+\frac{B}{(3n+1)(3n+4)}

    You'll find A=\frac{1}{6}, \ B=-\frac{1}{6}

    So, \frac{1}{(3n-2)(3n+1)(3n+4)}=\frac{1}{6}\left(\frac{1}{(3n-2)(3n+1)}-\frac{1}{(3n+1)(3n+4)}\right)

    Now apply the sum and see what happens.
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