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Thread: Binomial with 3 terms inside brackets

  1. #1
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    Binomial with 3 terms inside brackets

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

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

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

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

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

    $\displaystyle 2i = 4$

    $\displaystyle i = 2$

    $\displaystyle 5\choose{2}$$\displaystyle (1)^2$

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

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

    $\displaystyle (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 ]=$$\displaystyle \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 $\displaystyle x^4$, so:

    $\displaystyle 8-2k+i=4, \: k=\overline{0,4} \: and \: i=\overline{0,k}$. 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 $\displaystyle x^4$ in the series expansion of $\displaystyle (1+x+x^2)^{-4}$?

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

    Therefore $\displaystyle (1+x+x^2)^{-4} = (1-x)^4(1-x^3)^{-4} = (1-4x+6x^2-4x^3+x^4)(1+4x^3+\ldots)$, from which you can easily pick out the coefficient of $\displaystyle x^4.$
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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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