A) \[^{10}{{C}_{4}}\]
B) \[^{10}{{C}_{4}}{{.2}^{4}}x\]
C) \[{{2}^{4}}.{{x}^{2}}\]
D) \[^{10}{{C}_{4}}{{.2}^{4}}\left( \frac{1}{{{x}^{2}}} \right)\]
Correct Answer: D
Solution :
[d] The 7th term from the end = 5th term from beginning \[{{T}_{5}}{{=}^{10}}{{C}_{4}}{{x}^{6}}{{\left( -\frac{2}{{{x}^{2}}} \right)}^{4}}={{\,}^{10}}{{C}_{4}}{{.2}^{4}}\left( \frac{1}{{{x}^{2}}} \right)\]
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