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Math Help - Question regarding Binomial Coefficents.

  1. #1
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    Question regarding Binomial Coefficents.

    Could anyone help me solve me the following problem please :-

    Find the coefficent of x^27 in the binomial expansion of ( 3/x + x^2 )^18.

    I can work my way through the initial steps but do not know how to go on from there. Thanks in advance.
    Last edited by mr fantastic; September 19th 2010 at 09:17 PM. Reason: Edited title.
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  2. #2
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    Quote Originally Posted by aamiri View Post
    Could anyone help me solve me the following problem please :-

    Find the coefficent of x^27 in the binomial expansion of ( 3/x + x^2 )^18.

    I can work my way through the initial steps but do not know how to go on from there. Thanks in advance.

    By Newton's binomial Theorem, \left(\frac{3}{x}+x^2\right)^{18}=\sum\limits_{k=0  }^{18}\binom {18}{k}3^kx^{-k}x^{2(18-k)} =\sum\limits_{k=0}^{18}\binom {18}{k}3^kx^{36-3k}

    Since you want the coefficient of x^{27} , you need to know for which index k we get

    36-3k=27\Longleftrightarrow 3k=9\Longleftrightarrow k=3 , so the wanted coefficient is \binom{18}{3}3^3.

    Now do the maths and evaluate this number.

    tonio
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  3. #3
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    Hello, aamiri!

    \text{Find the coefficent of }x^{27}\text{ in the binomial expansion of }\left(\frac{3}{x} + x^2 \right)^{18}

    We have: . \left(3x^{-1} + x^2\right)^{18}

    The terms are of the form: . \displaystyle {18\choose a,b}\left(3x^{-1}\right)^a\left(x^2\right)^b \;\;\text{ where }a+b \,=\,18 .[1]


    The terms we want are of the form: . \left(x^{-1}\right)^a(x^2)^b \;=\;x^{27}

    . . Then: . (x^{-a})(x^{2b}) \:=\:x^{27} \quad\Rightarrow \quad x^{2b-a} \:=\:x^{27}


    \begin{array}{ccccc}\text{We have:} & 2b - a &=& 27 \\<br />
\text{Add [1]:} & b + a &=& 18 \end{array}

    Hence: . 3b \,=\,45 \quad\Rightarrow\quad b\,=\,15 \quad\Rightarrow\quad a\,=\,3


    It seems that there is only one term that contains x^{27}.

    That term is: . \displaystyle {18\choose3,15}\left(3x^{-1}\right)^3\left(x^2\right)^{15}

    . . . . . . . . =\;(816)\,\left(3^3x^{-3}\right)\left(x^{30}\right)

    . . . . . . . . =\;\boxed{22,\!032}\,x^{27}



    Edit: Ha! . . . This is rehash of tonio's solution . . . *blush*
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