Question

The $6^{\text{th}}$ term in the expansion of ${(2x^2-\frac{1}{3x^2})}^{10}$ is

• A. $-\frac{35840}{9}$
• B. $\frac{560}{243}$
• C. $-\frac{896}{27}$
• D. $-\frac{764}{25}$

Answer 1

The correct option is C $-\frac{896}{27}$

We know that, $(r+1{)}^{\text{th}}$ term in the expansion of $(x+y{)}^n$ is
$T_{r+1}={}^n{C}_r{x}^{n-r}{y}^r$
Hence for ${(2x^2-\frac{1}{3x^2})}^{10}$
$T_6={}^{10}{C}_5(2x^2{)}^5{(-\frac{1}{3x^2})}^5$
$=-\frac{10!}{5!\times 5!}\times 32\times \frac{1}{243}$

$=-\frac{896}{27}$