Question

Show that the expansion of ${\left(x^2+\frac{1}{x}\right)}^{12}$ does not contain any term involving x⁻¹.

Answer 1

Suppose x⁻¹ occurs at the (r + 1)th term in the given expression.
Then,
$$\begin{gathered} T_{r+1}={}^{12}C_r(x^2{)}^{12-r}{\left(\frac{1}{x}\right)}^r \\ ={}^{12}C_r{x}^{24-2r-r} \\ \mathrm{For}\mathrm{this}\mathrm{term}\mathrm{to}\mathrm{contain}{x}^{-1},\mathrm{we}\mathrm{must}\mathrm{have} \\ 24-3r=-1 \\ \Rightarrow 3r=25 \\ \Rightarrow r=\frac{25}{3} \\ \mathrm{It}\mathrm{is}\mathrm{not}\mathrm{possible},\mathrm{as}r\mathrm{is}\mathrm{not}\mathrm{an}\mathrm{integer}. \end{gathered}$$

Hence, the expansion of ${\left(x^2+\frac{1}{x}\right)}^{12}$ does not contain any term involving x⁻¹.