Question

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

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

Answer 1

The correct option is D ${}^{12}{C}_4{.2}^4$

$T_{r+1}={}^n{C}_r{a}^{n-r}{b}^r$
Applying to the above question we get
$T_{r+1}=(-1){}^r{}^{12}{C}_r\left(2x\right){}^{12-r}(2x^2{)}^{-r}$
$=(-1){}^r{}^{12}{C}_r\left(2\right){}^{12-2r}(x{)}^{12-3r}$
For the term independent of $x$, power of $x$ should be zero
$12-3r=0$
$r=4$
$T_5={}^{12}{C}_4\left(2\right){}^4$