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Math Help - partial fractions and binomial expansion

  1. #1
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    partial fractions and binomial expansion

    I have expressed the expression as partial fractions and I have done binomial expansion up to x cubed. The answer in the book has different coefficients of x squared and x cubed. I have had a good look at it. Can someone check it?
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  2. #2
    MHF Contributor Unknown008's Avatar
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    I got +1/2 as the numerator of 1/(2-x).

    Check this one again.

    EDIT: I took the wrong denominator, that is 1/(x-2). Problem resolved later.
    Last edited by Unknown008; October 27th 2010 at 10:33 AM.
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  3. #3
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    I can't find any mistake with the partial fractions. It checks ok when reversed.
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  4. #4
    MHF Contributor Unknown008's Avatar
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    Oh right. Sorry, I was using 1/(x-2)

    In your simplification line of the first term.

    \dfrac{-1.-2}{2!} \left(\dfrac{x}{4}\right)^2 \neq \dfrac{3x^2}{32}

    \dfrac{-1.-2}{2!} \left(\dfrac{x}{4}\right)^2 = \dfrac{x^2}{16}

    The binomial coefficient is not to add the numerator but to multiply them.
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  5. #5
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    Yes its a simpler mistake than I was expecting. I am only used to positive integral powers.
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