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Math Help - The coefficient of x^10 in this expansion

  1. #1
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    Talking The coefficient of x^10 in this expansion

    Hi, I've been doing Binomial Theorem, and here is an annoying question I encountered:

    Find the coefficient of x^{10} in the expansion of:
    \left ( x - \frac{1}{x} \right )^6 \left ( x+\frac{1}{x} \right )^8

    Thanks.
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    Quote Originally Posted by differentiate View Post
    Hi, I've been doing Binomial Theorem, and here is an annoying question I encountered:

    Find the coefficient of x^{10} in the expansion of:
    \left ( x - \frac{1}{x} \right )^6 \left ( x+\frac{1}{x} \right )^8

    Thanks.
    The general term is {6 \choose r} x^r \left(- \frac{1}{x} \right)^{6 - r} \cdot {8 \choose t} x^t \left(\frac{1}{x} \right)^{8 - t} = (-1)^{6 - r} {6 \choose r} {8 \choose t} x^{r + r - 6} x^{t + t - 8}

     = (-1)^{6 - r} {6 \choose r} {8 \choose t} x^{2r + 2t - 14}.

    You require the power of x to be 10. Therefore:

    2r + 2t - 14 = 10 \Rightarrow r + t = 12.

    Construct the required coefficient from the values of r and t that satisfy the above equation.
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    OMG you're a genius!! THANK YOU!!!!!!!!!!!!!!!!!!!!! Your method works much better than the one described in the textbook!!!!!!!!!!!!!!!!!!!!!!! YOU GENIUS!!
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    Quote Originally Posted by differentiate View Post
    OMG you're a genius!! THANK YOU!!!!!!!!!!!!!!!!!!!!! Your method works much better than the one described in the textbook!!!!!!!!!!!!!!!!!!!!!!! YOU GENIUS!!
    No doubt someone will post a better answer that will make me look somewhat less than a genius. (I was going to first simplify the expression using the difference of two squares formula but someone else might like to expand (ha ha) upon that).
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    Quote Originally Posted by mr fantastic View Post
    No doubt someone will post a better answer that will make me look somewhat less than a genius.
    No one would dare!

    (I was going to first simplify the expression using the difference of two squares formula but someone else might like to expand (ha ha) upon that).
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