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Math Help - Sum to Infinity of a Geometric Series

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    Sum to Infinity of a Geometric Series

    Q. Find in terms of x, in its simplest form, the sum of the infinite geometric series

    x^2 + x^2/ (1 - x) + x^2/ (1 - x)^2 + x^2/ (1 - x)^3 +

    Eq.: S = a/ 1 - r

    Ans.: From text-book, x(x - 1)

    Attempt: a = x^2, r = 1/ (1 - x)

    x^2/ (1 - (1/ (1 - x)))
    x^2(-1 + x) . inverting 1/ (1 - x) and changing +/ - via the 1 - from 1 - 1/(1 - x)
    x^3 - x^2
    x^2(x - 1)

    This is as close to the answer as I can get. I cant quite work it out and was hoping someone could help. Thank you.
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    Re: Sum to Infinity of a Geometric Series

    Quote Originally Posted by GrigOrig99 View Post
    Q. Find in terms of x, in its simplest form, the sum of the infinite geometric series

    x^2 + x^2/ (1 - x) + x^2/ (1 - x)^2 + x^2/ (1 - x)^3 +

    Eq.: S = a/ 1 - r

    Ans.: From text-book, x(x - 1)

    Attempt: a = x^2, r = 1/ (1 - x)

    x^2/ (1 - (1/ (1 - x)))
    x^2(-1 + x) . inverting 1/ (1 - x) and changing +/ - via the 1 - from 1 - 1/(1 - x)
    x^3 - x^2
    x^2(x - 1)

    This is as close to the answer as I can get. I cant quite work it out and was hoping someone could help. Thank you.
    Your a and r values are fine.

    If we look at the denominator (1-r) and simplify it: 1-\dfrac{1}{1-x} = \dfrac{1-x-1}{1-x} = \dfrac{-x}{1-x}

    This makes our sum S_{\infty} = \dfrac{x^2}{\left(\dfrac{-x}{1-x}\right)} = x^2 \cdot \dfrac{1-x}{-x} = -x(1-x) = x(x-1)
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    Re: Sum to Infinity of a Geometric Series

    Thanks for clearing that up.
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