Question

The constant terms is the expansion of $ {{\left( \sqrt{x}-\frac{c}{{{x}^{2}}} \right)}^{10}} $ is $ 180, $ then the value of $c$ equals to

• A. $ \pm \,\,2 $
• B. $ \pm \,\,3 $
• C. $ \pm \,\,4 $
• D. $None\, of \,these$

Answer 1

Answer: (A)

Given, $ {{\left( \sqrt{x}\,-\frac{C}{{{x}^{2}}} \right)}^{10}} $
General term is $ {{T}_{r+1}}{{=}^{10(}}{{C}_{r}}{{(\sqrt{x})}^{10-r}}{{\left( -\frac{C}{{{x}^{2}}} \right)}^{r}} $
$ {{=}^{10}}{{C}_{r}}\,{{x}^{(5-r/2)}}{{(-C)}^{r}}{{x}^{-2r}} $
$ {{=}^{10\,}}{{C}_{r}}\,{{x}^{(5-r/2-2r)}}\,{{(-C)}^{r}} $
$ {{=}^{10\,}}{{C}_{r}}\,{{x}^{\left( 5-\frac{5r}{2} \right)}}\,{{(-C)}^{r}} $
For the constant term, Put $ 5-\frac{5r}{2}=0 $
$ \Rightarrow $ $ (r=2) $
$ \Rightarrow $ $ {{T}_{2+1}}{{=}^{10}}{{C}_{2}}\,\,{{x}^{(5-5)}}\,{{(-C)}^{2}}{{=}^{10}}{{C}_{2}}.{{C}^{2}} $
Given, $ {{T}_{3}}=180{{=}^{10}}{{C}_{2}}.{{C}^{2}} $
$ \Rightarrow $ $ {{C}^{2}}.\frac{10\times 9}{2}=180 $
$ \Rightarrow $ $ {{C}^{2}}=4 $
$ \Rightarrow $ $ C=\pm 2 $