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Math Help - appeal to the geometric series

  1. #1
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    appeal to the geometric series

    Derive the indicated result by appealing to the geometric series

    \sum\limits_{k = 0}^\infty (-1)^kx^k =1/(1+x), |x| < 1

    \sum\limits_{k = 0}^\infty (-1)^kx^{2k} =1/(1+x^2), |x| < 1


    I dun understand what the question is asking, the geometric series is 1/(1-x), how does it turn into 1/(1+x)
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  2. #2
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    Quote Originally Posted by wopashui View Post
    Derive the indicated result by appealing to the geometric series

    \sum\limits_{k = 0}^\infty (-1)^kx^k =1/(1+x), |x| < 1

    \sum\limits_{k = 0}^\infty (-1)^kx^{2k} =1/(1+x^2), |x| < 1


    I dun understand what the question is asking, the geometric series is 1/(1-x), how does it turn into 1/(1+x)
    You should know that \sum\limits_{k = 0}^{\infty} r^k = \frac{1}{1 - r} for |r| < 1.

    In your first question r = -x and in your second question r = -x^2.
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  3. #3
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    Quote Originally Posted by wopashui View Post
    I dun understand what the question is asking, the geometric series is 1/(1-x), how does it turn into 1/(1+x)
    Simple. Just replace x by x.
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