Thank you for reporting, we will resolve it shortly
Q.
The coefficient of the term independent of $x$ in the expansion of $ {{\left( \sqrt{x}+\frac{1}{\sqrt{x}} \right)}^{10}} $ is equal to
J & K CETJ & K CET 2012Binomial Theorem
Solution:
$ \therefore $ The general term in the expansion of
$ \left( \sqrt{x}+\frac{1}{\sqrt{x}} \right) $ is $ {{T}_{r+1}}{{=}^{10}}{{C}_{r}}{{(\sqrt{x})}^{10-r}}{{\left( \frac{1}{\sqrt{x}} \right)}^{r}} $
$ {{=}^{10}}{{C}_{r}}{{x}^{\frac{10-r}{2}}}\left( {{x}^{-\frac{r}{2}}} \right) $
$ {{=}^{10}}{{C}_{r}}{{x}^{\frac{10-2r}{2}}} $
For independent of x, Put
$ \frac{10-2r}{2}=0, $ we get $ r=5 $ s
$ \therefore $ Required coefficient $ {{=}^{10}}{{C}_{5}} $
$ =\frac{10\times 9\times 8\times 7\times 6}{5\times 4\times 3\times 2\times 1} $ $ =252 $