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Math Help - [SOLVED] Series - Geometric Sum Formula

  1. #1
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    [SOLVED] Series - Geometric Sum Formula

    Problem:
    Find a formula for
    a.
    \frac{1}{1+x^2}+\frac{1}{(1+x^2)^2}+...+\frac{1}{(  1+x^2)^n}
    b. Also, show that if a \neq 0 , then
    \frac{1}{a}=1+(1-a)+(1-a)^2+\frac{(1-a)^3}{a}
    ========================================
    a. By the Geometric Sum Formula,
    \sum_{k=0}^{n}r^k  = \frac{1-r^{n+1}}{1-r} (1)

    Since \frac{1}{1+x^2}+\frac{1}{(1+x^2)^2}+...+\frac{1}{(  1+x^2)^n}= \sum_{k=0}^{n} \left( \frac{1}{1+x^2} \right)^n = \sum_{k=0}^{n} \frac{1}{(1+x^2)^n} (2)

    Let r= \frac{1}{1+x^2}, then substitute r back into (1) yields

    \frac{1-\left( \frac{1}{1+x^2} \right)^{n+1}}{1- \frac{1}{1+x^2}} (3)

    Before I simplify this fraction, this formula doesn't seem right to me.
    Suppose n=2 and x=2, then the equation (2) becomes

    \frac{1}{1+2^2}+\frac{1}{(1+2^2)^2}= \frac{1}{5} + \frac{1}{25} = \frac{6}{25}

    However, if you plug n =2 and x = 2 into equation (3) , then it becomes
    \frac{1- \left( \frac{1}{5}\right)^3}{1-\frac{1}{5}} = \frac{1 - \frac{1}{125}}{1-\frac{1}{5}} = \frac{31}{25}

    But \frac{6}{25} \neq \frac{31}{25}

    (b) Since 1+r+r^2+...+r^n= \frac{1-r^{n+1}}{1-r}

    \implies [ 1 + (1-a) + (1-a)^2 ] + \frac{(1-a)^3}{a}

    \implies \frac{1-(1-a)^3}{1-(1-a)} + \frac{(1-a)^3}{a}

    \implies \frac{1-(1-a)^3}{a} + \frac{(1-a)^3}{a}
    = \frac{1}{a}

    I'm unsure what I am doing incorrectly for part (a).

    Thank you for your time.
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  2. #2
    Senior Member JaneBennet's Avatar
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    The formula for the first part should be

    \dfrac{\dfrac{1}{1+x^2}\left[1-\left(\dfrac{1}{1+x^2}\right)^n\right]}{1-\dfrac{1}{1+x^2}}
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  3. #3
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    Quote Originally Posted by JaneBennet View Post
    The formula for the first part should be

    \dfrac{\dfrac{1}{1+x^2}\left[1-\left(\dfrac{1}{1+x^2}\right)^n\right]}{1-\dfrac{1}{1+x^2}}
    Thank you Jane, this is because \sum_{k=0}^{n} \frac{1}{(1+x^2)^n}
    k actually starts at 1.

    After some algebra, I get \frac{1}{x^2} \left( 1 - \frac{1}{(1-x^2)^n}\right)
    Last edited by Paperwings; August 6th 2008 at 10:46 AM.
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