Question

The term independent of x in the expansion of $(\frac{2}{x}-\frac{1}{2x^2}{)}^{12}$

• A. ${}^{-12}{C}_3{2}^6$
• B. ${}^{-12}{C}_5{2}^2$
• C. ${}^{12}{C}_6$
• D. ${}^{-12}{C}_4{2}^4$

Answer 1

Answer: (D)

$T_{r+1}={}^{12}{c}_r{\left(2x\right)}^{12-r}{\left(-\frac{1}{2x^2}\right)}^r$
$={}^{12}{c}_r{\left(2\right)}^{12-r}{x}^{12-r-2r}{\left(-\frac{1}{2}\right)}^r$
$\therefore 12-3r=0$
$\Rightarrow 3r=12$
$\Rightarrow r=4$
$\therefore$ reqd. term $={}^{12}{c}_4{\left(2\right)}^8{\left(-\frac{1}{2}\right)}^4$
$={}^{12}{c}_4{2}^4$.