Question

Find the coefficient of $x^{10}$ and $x^9$ in the expansion of ${(2x^2-\frac{1}{x})}^{20}$

Answer 1

$T_{r+1}={}^{20}{C}_r\left(2x^2{)}^{20-r}\right(-1/x{)}^r$
$=(-1{)}^r{}^{20}{C}_r{2}^{20-r}{x}^{40-2r-r}$
$\therefore 40-3r=10$ or 9
$\therefore 3r=30$ or 31.
$\therefore$ $r=10$ the other value does not give integral value of r so that there will be term of $x^9$ .
Putting $r=10$
$T_{11}=(-1{)}^{10}{}^{20}{C}_{10}{2}^{20-10}{x}^{40-30}$
$=\frac{\left(20\right)!}{\left(10\right)!\left(10\right)!}{2}^{10}{x}^{10}$
Hence the required coefficient is $\frac{\left(20\right)!}{\left(10\right)!\left(10\right)!}\cdot 2^{10}$.