# Binomial Theorem

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• Dec 22nd 2008, 02:14 AM
acc100jt
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?
• Dec 22nd 2008, 03:28 AM
Skalkaz
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?
• Dec 22nd 2008, 04:36 AM
Plato
If you have not written this incorrectly, you should see that $\left( {\frac{{2x\sqrt x - 1}}
{{\sqrt x }}} \right) = 2x - x^{ - \frac{1}{2}}$

$\left( {2x - x^{ - \frac{1}
{2}} } \right)^n = \sum\limits_{k = 0}^n {\left( \begin{gathered}
n \hfill \\
k \hfill \\
\end{gathered} \right)\left( {2x} \right)^{n - k} \left( { - x^{ - \frac{1}
{2}} } \right)^k } = \sum\limits_{k = 0}^n {\left( { - 1} \right)^k \left( 2 \right)^{n - k} \left( \begin{gathered}
n \hfill \\
k \hfill \\
\end{gathered} \right)x^{n - \frac{{3k}}
{2}} }$