Question

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

• A. $ ^{-12} C_32^6 $
• B. $ ^{-12} C_52^2 $
• C. $^{12} C_6$
• D. $ ^{-12} C_42^4 $

Answer 1

Answer: (D)

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