Question

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

Answer 1

(a) 7920

$$\begin{gathered} \mathrm{Suppose}\mathrm{the}(r+1)\mathrm{th}\mathrm{term}\mathrm{is}\mathrm{independent}\mathrm{of}x\mathit{}\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{expansion}. \\ \mathrm{Then},\mathrm{we}\mathrm{have}: \\ {T}_{r+1}={}^{12}C_r(2x^2{)}^{12-r}{\left(\frac{-1}{x}\right)}^r \\ =(-1{)}^r{}^{12}C_r{2}^{12-r}{x}^{24-2r-r} \\ \mathrm{For}\mathrm{this}\mathrm{term}\mathrm{to}\mathrm{be}\mathrm{independent}\mathrm{of}x,\mathrm{we}\mathrm{must}\mathrm{have}: \\ 24-3r=0 \\ \Rightarrow r=8 \\ \therefore \mathrm{Required}\mathrm{term}=(-1{)}^8{}^{12}C_8{2}^{12-8}=\frac{12\times 11\times 10\times 9}{4\times 3\times 2}\times 2^4=7920 \end{gathered}$$