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Math Help - Binomial theorem help

  1. #1
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    Binomial theorem help

    hey guys, question:
    Find the coefficient of x^{-4} in (4x^7-2x^{-3})^8

    I know we are ment to rewrite (4x^7-2x^{-3})^8 so there are two expressions to the power of 8 but im not sure how we are ment to rewrite it. I got the answer (x^{-3})^8 (4x^{10} -2)^8 but not sure how they got this... then it says we see that finding the coefficient of x^{-4} is equivalent to finding the coefficient of x^{20}.. how did they get x^{20} ?

    thank you..
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  2. #2
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    Quote Originally Posted by jvignacio View Post
    hey guys, question:
    Find the coefficient of x^{-4} in (4x^7-2x^{-3})^8

    I know we are ment to rewrite (4x^7-2x^{-3})^8 so there are two expressions to the power of 8 but im not sure how we are ment to rewrite it. I got the answer (x^{-3})^8 (4x^{10} -2)^8 but not sure how they got this... then it says we see that finding the coefficient of x^{-4} is equivalent to finding the coefficient of x^{20}.. how did they get x^{20} ?

    thank you..
    The general term is {8 \choose r} (4x^7)^{8 - r} \left(-\frac{2}{x^3} \right)^r = {8 \choose r} 4^{8-r} (-2)^r  x^{56-7r - 3r}.

    Find the value of r such that 56 - 7r - 3r = -4 and use it ot get the coefficient.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    The general term is {8 \choose r} (4x^7)^{8 - r} \left(-\frac{2}{x^3} \right)^r = {8 \choose r} 4^{8-r} (-2)^r x^{56-7r - 3r}.

    Find the value of r such that 56 - 7r - 3r = -4 and use it ot get the coefficient.
    wow that totally makes more sense...! thanks mate
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