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Thread: Binomial theorem question

  1. #1
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    Binomial theorem question

    If n≥8, and the coefficients of $\displaystyle x^7$ and $\displaystyle x^8$ in the expansion of $\displaystyle (3+x/2)^n$ are equal, what is n?


    I equated the coefficents, but end up with n being 8.03 something, which can't be right.

    Any help?
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  2. #2
    Senior Member Peritus's Avatar
    Joined
    Nov 2007
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    $\displaystyle
    \left( {3 + \frac{x}
    {2}} \right)^n = \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}c}
    n \\
    k \\

    \end{array} } \right)3^k \left( {\frac{x}
    {2}} \right)^{n - k} }
    $

    now we equate the aforementioned coefficients:

    $\displaystyle
    \left( {\begin{array}{*{20}c}
    n \\
    {n - 7} \\

    \end{array} } \right)3^{n - 7} \left( {\frac{1}
    {2}} \right)^7 = \left( {\begin{array}{*{20}c}
    n \\
    {n - 8} \\

    \end{array} } \right)3^{n - 8} \left( {\frac{1}
    {2}} \right)^8
    $


    $\displaystyle \begin{gathered}
    \Leftrightarrow 6\frac{{n!}}
    {{7!\left( {n - 7} \right)!}} = \frac{{n!}}
    {{8!\left( {n - 8} \right)!}} \hfill \\
    \Leftrightarrow 48 = n - 7 \Rightarrow n = 55 \hfill \\
    \end{gathered} $
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