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Thread: Binomial Theorem

  1. #1
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    Binomial Theorem

    The n-th term of the expression $\displaystyle \left( \frac{2x\sqrt{x}-1}{\sqrt{x}}\right)^{n}$ when expanded in descending powers of $\displaystyle x$ is independent of $\displaystyle x$. Find the value of this term.

    Any idea?
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  2. #2
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    Dear acc100jt,

    the basic (how says man the "not exponent") is 2x-x^(1/2).

    And the formula is
    $\displaystyle (a + b)^n = \sum \frac{n!}{m!(n-m)!} a^m*b^{n-m}$
    index m run from 0 to n.

    For which one m is the term independent from x?
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  3. #3
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    If you have not written this incorrectly, you should see that $\displaystyle \left( {\frac{{2x\sqrt x - 1}}
    {{\sqrt x }}} \right) = 2x - x^{ - \frac{1}{2}} $
    $\displaystyle \left( {2x - x^{ - \frac{1}
    {2}} } \right)^n = \sum\limits_{k = 0}^n {\left( \begin{gathered}
    n \hfill \\
    k \hfill \\
    \end{gathered} \right)\left( {2x} \right)^{n - k} \left( { - x^{ - \frac{1}
    {2}} } \right)^k } = \sum\limits_{k = 0}^n {\left( { - 1} \right)^k \left( 2 \right)^{n - k} \left( \begin{gathered}
    n \hfill \\
    k \hfill \\
    \end{gathered} \right)x^{n - \frac{{3k}}
    {2}} } $
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