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Math Help - Binomial Coefficient

  1. #1
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    Binomial Coefficient

    Find the coefficient of x^5 in the expansion of (x^2 - 3/x)^7.
    Hi, how do you solve this question using binomial theorem. Thanks.
    Please help
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  2. #2
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    Ever term in the expansion of \left( {x^2  - 3x^{ - 1} } \right)^7 has the form C\left( {x^2 } \right)^j \left( { - 3x^{ - 1} } \right)^k ,\;j + k = 7\;\& \quad C = {7 \choose j}.

    So \left\{ \begin{array}{r}<br />
 j + k = 7 \\ <br />
 2j - k = 5 \\ <br />
 \end{array} \right.
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  3. #3
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    What happens next? As I can't figure these questions out at all.
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  4. #4
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    Quote Originally Posted by haku View Post
    What happens next? As I can't figure these questions out at all.
    You solve of j & k.
    Use those values to write the term.
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  5. #5
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    Quote Originally Posted by haku View Post
    Find the coefficient of x^5 in the expansion of (x^2 - 3/x)^7.
    Hi, how do you solve this question using binomial theorem. Thanks.
    Please help
    For the x's only:
    1st term, (x^2)^7 = x^14
    2nd term, [(x^2)^6][1/x] = x^11
    3rd term, [(x^2)^5][1/(x^2)] = x^8
    4th term, [(x^2)^4][1/(x^3)] = x^5 -------so it is in the 4th term.

    The whole 4th term has a binomial coefficent of
    nCr = 7C4
    = 7! / [(7-4)! *4!]
    = 7! / [3! *4!]
    = 7*6*5 / 3*2*1
    = 35

    So, the whole 4th term is
    35[(x^2)^4][(-3/x)^3]
    = 35(-27)x^5
    = -945x^5

    Therefore, the coefficient is -945. -------answer.
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