A) \[\frac{504}{259}\]
B) \[\frac{450}{263}\]
C) \[\frac{405}{256}\]
D) None of these
Correct Answer: C
Solution :
\[nth\]term in the expansion of \[{{\left( \frac{x}{2}-\frac{3}{{{x}^{2}}} \right)}^{10}}\] \[{{T}_{n+1}}{{=}^{10}}{{C}_{n}}{{\left( \frac{x}{2} \right)}^{10-n}}{{\left( -\frac{3}{{{x}^{2}}} \right)}^{n}}\] \[{{=}^{10}}{{C}_{n}}{{\left( \frac{1}{2} \right)}^{10-n}}{{(-1)}^{n}}{{3}^{n}}.{{x}^{10-n-2n}}\] \[={{(-1)}^{n}}^{10}{{C}_{n}}{{\left( \frac{1}{2} \right)}^{10-n}}{{.3}^{n}}.{{x}^{10-3n}}\] For the coefficient of \[{{x}^{4}},\] \[10-3n=4\] \[\Rightarrow \] \[3n=6\] \[\Rightarrow \] \[n=2\] \[\therefore \]Coefficient of\[{{x}^{4}}={{(-1)}^{2}}^{10}{{C}_{2}}{{\left( \frac{1}{2} \right)}^{8}}{{.3}^{2}}\] \[=\frac{10\times 6}{2\times 1}.\frac{9}{256}\] \[=\frac{5\times 81}{256}\] \[=\frac{405}{256}\]
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