Question

$6^{th}$ term in expansion of ${\left(2x^2-\frac{1}{3x^2}\right)}^{10}$ is:

• A. $\frac{4580}{17}$
• B. $-\frac{896}{27}$
• C. $\frac{5580}{17}$
• D. none of these

Answer 1

Answer: (B)

The general term in the expansion of $(x+a{)}^n$ is
$T_{r+1}={}^n{C}_r{x}^{n-r}{d}^r$
Given expansion is ${\left(2x^2-\frac{1}{3x^2}\right)}^{10}$
$\therefore T_{r+1}={}^{10}{C}_r{\left(2x^2\right)}^{10-r}{\left(-\frac{1}{3x^2}\right)}^r$
$\Rightarrow T_6={}^{10}{C}_5{\left(2x^2\right)}^5{\left(-\frac{1}{3x^2}\right)}^5$
$=-\frac{10!\times 32}{5!5!}\times \frac{1}{243}=-\frac{896}{27}$