Question

If rth term in the expansion of ${\left(2x^2-\frac{1}{x}\right)}^{12}$ is without x, then r is equal to
(a) 8
(b) 7
(c) 9
(d) 10

Answer 1

(c) 9
$$\begin{gathered} r\mathrm{th}\mathrm{term}\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{expansion}\mathrm{is}{}^{12}C_{r-1}(2x^2{)}^{12-r+1}{\left(\frac{-1}{x}\right)}^{r-1} \\ =(-1{)}^{r-1}{}^{12}C_{r-1}{2}^{13-r}{x}^{26-2r-r+1} \\ Forthistermtobeindependentofx,wemusthave: \\ 27-3r=0 \\ \Rightarrow r=9 \\ \mathrm{Hence},\mathrm{the}9\mathrm{th}\mathrm{term}\mathrm{in}\mathrm{the}\mathrm{expansion}\mathrm{is}\mathrm{independent}\mathrm{of}\mathrm{x}. \end{gathered}$$