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Math Help - Binomial with 3 terms inside brackets

  1. #1
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    Binomial with 3 terms inside brackets

    What is the coefficient of x^4 in the series expansion of (1+x+x^2)^{-4}?

    This would be easy if it were just (1+x^2)^{-4};however, I don't know what to do with the three terms.

    Normally I would find the coefficient of such a problem as follows:

    x^4 in (1+x^2)^{-4}

    = -4\choose{i} (x^2)^i

    2i = 4

    i = 2

    5\choose{2} (1)^2

    = 10
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  2. #2
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    Re: Binomial with 3 terms inside brackets

    (1+x+x^2)^{-4}=\frac{1}{(1+x+x^2)^4}

    (1+x+x^2)^4=[x^2+(1+x)]^4=\sum_{k=0}^{4}\binom{4}{k}(x^2)^{4-k}(1+x)^{k}=\sum_{k=0}^{4}\left [\binom{4}{k}x^{8-2k}\sum_{i=0}^{k}\binom{k}{i}1^{k-i}x^i\right ]= \sum_{k=0}^{4}\left [\binom{4}{k}x^{8-2k}\sum_{i=0}^{k}\binom{k}{i}x^i\right ]

    You need the coefficient of x^4, so:

    8-2k+i=4, \: k=\overline{0,4} \: and \: i=\overline{0,k}. Find k and i ^^
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  3. #3
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    Re: Binomial with 3 terms inside brackets

    Quote Originally Posted by terrorsquid View Post
    What is the coefficient of x^4 in the series expansion of (1+x+x^2)^{-4}?

    This would be easy if it were just (1+x^2)^{-4};however, I don't know what to do with the three terms.
    1+x+x^2 = \frac{1-x^3}{1-x} (sum of geometric series).

    Therefore (1+x+x^2)^{-4} = (1-x)^4(1-x^3)^{-4} = (1-4x+6x^2-4x^3+x^4)(1+4x^3+\ldots), from which you can easily pick out the coefficient of x^4.
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  4. #4
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    Re: Binomial with 3 terms inside brackets

    By the way, because there are three terms, this is a "tri"-nomial, not a "bi"-nomial?
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