1. ## 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?

2. 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?

3. If you have not written this incorrectly, you should see that $\displaystyle \left( {\frac{{2x\sqrt x - 1}} {{\sqrt x }}} \right) = 2x - x^{ - \frac{1}{2}}$
$\displaystyle \left( {2x - x^{ - \frac{1} {2}} } \right)^n = \sum\limits_{k = 0}^n {\left( \begin{gathered} n \hfill \\ k \hfill \\ \end{gathered} \right)\left( {2x} \right)^{n - k} \left( { - x^{ - \frac{1} {2}} } \right)^k } = \sum\limits_{k = 0}^n {\left( { - 1} \right)^k \left( 2 \right)^{n - k} \left( \begin{gathered} n \hfill \\ k \hfill \\ \end{gathered} \right)x^{n - \frac{{3k}} {2}} }$