Question

The term without x in the expansion of ${\left(2x-\frac{1}{2x^2}\right)}^{12}$ is
(a) 495
(b) −495
(c) −7920
(d) 7920

Answer 1

(d) 7920

$$\begin{gathered} \mathrm{Suppose}\mathrm{the}(\mathrm{r}+1)\mathrm{th}\mathrm{term}\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{expansion}\mathrm{is}\mathrm{independent}\mathrm{of}x. \\ \mathrm{Then},\mathrm{we}\mathrm{have}: \\ {T}_{r+1}={}^{12}C_r(2x{)}^{12-r}{\left(\frac{-1}{2x^2}\right)}^r \\ =(-1{)}^r{}^{12}C_r{2}^{12-2r}{x}^{12-r-2r} \\ \mathrm{For}\mathrm{this}\mathrm{term}\mathrm{to}\mathrm{be}\mathrm{independent}\mathrm{of}x,\mathrm{we}\mathrm{must}\mathrm{have}: \\ 12-3r=0 \\ \Rightarrow r=4 \\ \therefore \mathrm{Required}\mathrm{term}: \\ (-1{)}^4{}^{12}C_4{2}^{12-8} \\ =\frac{12\times 11\times 10\times 9}{4\times 3\times 2}\times 16 \\ =7920 \end{gathered}$$