A) \[\frac{160}{9}\]
B) \[\frac{80}{9}\]
C) \[\frac{160}{27}\]
D) \[\frac{80}{3}\]
Correct Answer: C
Solution :
\[1(6-r)+(-1)r=0\Rightarrow r=3,\] therefore fourth term will be independent of x i.e. \[^{6}{{C}_{3}}{{(2x)}^{3}}{{\left( \frac{1}{3x} \right)}^{3}}=20\times 8\times \frac{1}{27}=\frac{160}{27}\]You need to login to perform this action.
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