A) \[\frac{4580}{17}\]
B) \[-\frac{896}{27}\]
C) \[\frac{5580}{17}\]
D) None of these
Correct Answer: B
Solution :
Applying \[{{T}_{r+1}}={{\,}^{n}}{{C}_{r}}{{x}^{n-r}}{{a}^{r}}\]for \[{{(x+a)}^{n}}\] Hence \[{{T}_{6}}={{\,}^{10}}{{C}_{5}}{{(2{{x}^{2}})}^{5}}{{\left( -\frac{1}{3{{x}^{2}}} \right)}^{5}}\] \[=-\frac{10\,!}{5\,!\,5\,!}32\times \frac{1}{243}=-\frac{896}{27}\]You need to login to perform this action.
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