Question

The term independent of $x$ in the expansion of $(\frac{1}{60}-\frac{x^8}{81})\cdot {(2x^2-\frac{3}{x^2})}^6$ is equal to:

• A. $-108$
• B. $-72$
• C. $-36$
• D. $36$

Answer 1

The correct option is C $-36$

Term independent of $x$ in the expansion is
$\frac{1}{60}\times$ term independent of $x\text{in}{(2x^2-\frac{3}{x^2})}^6-\frac{1}{81}\times$ term with coefficient $x^{-8}\text{in}{(2x^2-\frac{3}{x^2})}^6$

Term independent of $x$ in ${(2x^2-\frac{3}{x^2})}^6$
$\because T_{r+1}=6C_r(2x^2{)}^{6-r}\times {\left(\frac{-3}{x^2}\right)}^r$
$\Rightarrow 2(6-r)+(-2)r=0\Rightarrow r=3...\left(1\right)$
and term with coefficient $x^{-8}\text{in}{(2x^2-\frac{3}{x^2})}^6$
$\because T_{r+1}=6C_r(2x^2{)}^{6-r}\times {\left(\frac{-3}{x^2}\right)}^r$
$\Rightarrow 2(6-r)+(-2)r=-8\Rightarrow r=5...\left(2\right)$
From $\left(1\right)$ and $\left(2\right)$
The term independent of $x$ in the given expansion is -
$=\frac{1}{60}\times {}^6{C}_3\times 2^3\times (-3{)}^3+(-\frac{1}{81})\times {}^6{C}_5\times 2\times (-3{)}^5=-72+36=-36$