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Math Help - Binomial Theorem

  1. #1
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    Binomial Theorem

    The n-th term of the expression \left( \frac{2x\sqrt{x}-1}{\sqrt{x}}\right)^{n} when expanded in descending powers of x is independent of 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
    (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 \left( {\frac{{2x\sqrt x  - 1}}<br />
{{\sqrt x }}} \right) = 2x - x^{ - \frac{1}{2}}
    \left( {2x - x^{ - \frac{1}<br />
{2}} } \right)^n  = \sum\limits_{k = 0}^n {\left( \begin{gathered}<br />
  n \hfill \\<br />
  k \hfill \\ <br />
\end{gathered}  \right)\left( {2x} \right)^{n - k} \left( { - x^{ - \frac{1}<br />
{2}} } \right)^k }  = \sum\limits_{k = 0}^n {\left( { - 1} \right)^k \left( 2 \right)^{n - k} \left( \begin{gathered}<br />
  n \hfill \\<br />
  k \hfill \\ <br />
\end{gathered}  \right)x^{n - \frac{{3k}}<br />
{2}} }
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