Question

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

Answer 1

General term is

$T_r{=}^n{C}_r{x}^r{a}^{n-r}$

in $(x+a)r_n\left(rϵz\right)$

$\therefore T_r{=}^{12}{C}_r(x^2{)}^r{\left(\frac{1}{x}\right)}^{12-r}$

$\Rightarrow T_r{=}^{12}{C}_r{x}^{3r-12}$

for $x^{-1}$ term $3r-12=-1$

$\Rightarrow r=\frac{11}{3}$ not possible