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Math Help - Method of difference and recurrence relations

  1. #1
    Member helloying's Avatar
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    Jul 2008
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    Method of difference and recurrence relations

    I got some same type of qns that i don't know how to solve.

    Find: \sum^{n}_{r=1} \frac{1}{(2n+3)(2n+1)}
    I found and checked the ans to be \frac{1}{6} - \frac{1}{4n+1}
    Hence solve \sum^{n}_{r=1} \frac{1}{4n^2 -1}

    And another similar one,find \sum^{n}_{r=3} \frac{4-r}{r(r-1)(r-2)}
    hence find \sum^{n}_{r=1} \frac{2-r}{r(r+1)(r+2)}

    I dont know how to do the hence part. Btw sorry about the summation. i cant figure out how to use latex properly

    Thank you for ur time
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  2. #2
    MHF Contributor red_dog's Avatar
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    S_1=\displaystyle\sum_{r=1}^n\frac{1}{(2r+1)(2r+3)  }=\dfrac{1}{3\cdot 5}+\dfrac{1}{5\cdot 7}+\ldots+\dfrac{1}{(2n-1)(2n+1)}+\dfrac{1}{(2n+1)(2n+3)}

    S_2=\displaystyle\sum_{r=1}^n\dfrac{1}{4r^2-1}=\displaystyle\sum_{r=1}^n\dfrac{1}{(2r-1)(2r+1)}=

    =\dfrac{1}{1\cdot 3}+\dfrac{1}{3\cdot 5}+\ldots+\dfrac{1}{(2n-1)(2n+1)}=

    =\dfrac{1}{1\cdot 3}+S_1-\dfrac{1}{(2n+1)(2n+3)}

    Now replace S_1.

    You can continue in the same way for the rest of the problem.
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