A) \[\frac{504}{259}\]
B) \[\frac{450}{263}\]
C) \[\frac{405}{256}\]
D) none of these
Correct Answer: C
Solution :
Key Idea: General term in the expansion of\[{{(x+a)}^{n}}\]is \[{{T}_{r+1}}{{=}^{n}}{{C}_{r}}{{x}^{n}}{{a}^{n-r}}\] The general term of\[{{\left( \frac{x}{2}-\frac{3}{{{x}^{2}}} \right)}^{10}}\]is \[{{T}_{r+1}}{{=}^{10}}{{C}_{r}}{{\left( \frac{x}{2} \right)}^{r}}.{{\left( \frac{-3}{{{x}^{2}}} \right)}^{10-r}}\] \[{{=}^{10}}{{C}_{r}}{{\left( \frac{1}{2} \right)}^{r}}{{(-3)}^{10-r}}.{{x}^{r}}.{{x}^{-2(10-r)}}\] \[{{=}^{10}}{{C}_{r}}{{\left( \frac{1}{2} \right)}^{r}}{{(-3)}^{10-r}}{{x}^{r}}{{x}^{r-20+2r}}\] For coefficient of \[{{x}^{4}}\] \[r-20+2r=4\] \[\Rightarrow \] \[3r=24\] \[\Rightarrow \] \[r=8\] \[\therefore \] \[{{T}_{8+1}}{{=}^{10}}{{C}_{8}}{{\left( \frac{1}{2} \right)}^{8}}{{(-3)}^{10-8}}{{x}^{4}}\] \[{{=}^{10}}{{C}_{8}}{{\left( \frac{1}{2} \right)}^{8}}{{3}^{2}}.{{x}^{4}}\] \[\therefore \]Its coefficient is \[^{10}{{C}_{8}}{{(2)}^{-8}}{{3}^{2}}\] \[=\frac{405}{256}\] Note: The problem can be solved by taking general term as \[{{T}_{r+1}}{{=}^{10}}{{C}_{r}}{{\left( \frac{x}{2} \right)}^{10-r}}{{\left( \frac{-3}{{{x}^{2}}} \right)}^{r}}\]also.
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