Question

Find the coefficient of $x^2$ in ${(x^2+\frac{1}{x^3})}^6$.

Answer 1

Let $T_{r+1}$ be the term containing $x^2$.
$T_{r+1}{=}^6{C}_r(x^2{)}^{6-r}{\left(\frac{1}{x^3}\right)}^r{=}^6{C}_r{x}^{12-2r}\frac{1}{x^{3r}}{=}^6{C}_r{x}^{12-5r}$
As the coefficient of x is 2
$12-5r=2⟹r=2$.
$\therefore$ Coefficient of $x^2{=}^6{C}_2=15$.