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  1. #1
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    Urgent please help

    Find the coefficient of x4 in the expansion of (3x - 1/x)6
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  2. #2
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    \displaystyle\left(3x-\frac{1}{x}\right)^6

    The general term is T_{k+1}=(-1)^kC_6^k(3x)^{6-k}\left(\frac{1}{x}\right)^k=(-1)^kC_6^k3^{6-k}x^{6-2k}

    x^{6-2k}=x^4\Rightarrow 6-2k=4\Rightarrow k=1
    So, the second term is T_2=-6\cdot 3^5\cdot x^4=-1458x^4
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  3. #3
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    Note that there is certainly a MUCH shorter way doing this. However, for good pedagogy, I worked this expansion out completely for you (so you understand each step). I will assume you know how to compute binomial coefficients (i.e. 6 choose 4).

    Find the coefficient of x4 in the expansion of (3x - 1/x)6

    We want to expand  (3x - x^{-1})^6 .

    The Binomial Theorem states:

     (a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k

    We let:

     a = 3x

     b = -x^{-1}

     n = 6

    So,  (3x - x^{-1})^6 = \sum_{k=0}^6 \binom{6}{k} (3x)^{6-k} (-x^{-1})^k


      =  \binom{6}{0}(3x)^6 (-x^{-1})^0 + \binom{6}{1}(3x)^5 (-x^{-1})^1 + \binom{6}{2}(3x)^4 (-x^{-1})^2 + \binom{6}{3}(3x)^3  (-x^{-1})^3 + \binom{6}{4}(3x)^2 (-x^{-1})^4 + \binom{6}{5}(3x)^1 (-x^{-1})^5 + \binom{6}{6}(3x)^0 (-x^{-1})^6


      =  \binom{6}{0}729x^6 - \binom{6}{1}243x^5 x^{-1} + \binom{6}{2}81x^4 x^{-2} - \binom{6}{3}27x^3  x^{-3} + \binom{6}{4}9x^2 x^{-4} - \binom{6}{5}3x x^{-5} + \binom{6}{6} 1^{-6}


      =  \binom{6}{0}729x^6 - \binom{6}{1}243x^4 + \binom{6}{2}81x^2 - \binom{6}{3}27  + \binom{6}{4}9x^{-2} - \binom{6}{5}3x^{-4} + \binom{6}{6} 1^{-6}


     =  1(729x^6) - 6(243x^4) + 15(81x^2) - 20(27) + 15(9x^{-2}) - 6(3x^{-4}) + 1(1^{-6})


     = 729x^6 - 1458x^4 + 1215x^2 - 540 + \frac{135}{x^2} - \frac{18}{x^4} + \frac{1}{x^6}

    So, what does the coefficient of x^4 look like to you?

    -Andy
    Last edited by abender; July 29th 2008 at 03:45 PM.
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