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Math Help - Sum of Infinite Series

  1. #1
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    Sum of Infinite Series

    Just came across this problem in one of my past papers (which annoyingly dont come with answers) :

    Find as a function of x the sum of the series

    x + 3x^2 + 9x^3 + 27x^4 + ((-1)^n)*3^n*x^(n+1) + ....

    I have solved similar problems before with binomial and geometric series expressions but I'm not sure how I'd go about this particular series.

    Any help?
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  2. #2
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    Quote Originally Posted by thepars View Post
    Just came across this problem in one of my past papers (which annoyingly dont come with answers) :

    Find as a function of x the sum of the series

    x + 3x^2 + 9x^3 + 27x^4 + ((-1)^n)*3^n*x^(n+1) + ....

    I have solved similar problems before with binomial and geometric series expressions but I'm not sure how I'd go about this particular series.

    Any help?
    I think this is what you mean you have a (-1)^n in your general term but you sign does not alternate

    \sum_{n=0}^{\infty}(-1)^n3^{n}x^{n+1} =x\cdot \sum_{n=0}^{\infty}(-1)^n3^{n}x^{n}=x\cdot \sum_{n=0}^{\infty}(-3x)^n

    This is a geometric series so we get

    x\cdot \frac{1}{1+3x}=\frac{x}{1+3x}
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  3. #3
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    Quote Originally Posted by TheEmptySet View Post
    I think this is what you mean you have a (-1)^n in your general term but you sign does not alternate

    \sum_{n=0}^{\infty}(-1)^n3^{n}x^{n+1} =x\cdot \sum_{n=0}^{\infty}(-1)^n3^{n}x^{n}=x\cdot \sum_{n=0}^{\infty}(-3x)^n

    This is a geometric series so we get

    x\cdot \frac{1}{1+3x}=\frac{x}{1+3x}
    Wow it was so obvious I think I'm just having a bad day maths-wise.

    Thanks
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