Question

Find the constant term in expansion of
${(x-\frac{3}{x^2})}^9$

Answer 1

As we know that the general term of the expansion is given as-

${(a+b)}^{15}$ is given as-

$T_{r+1}={}^n{C}_r{a}^r{b}^{n-r}$

Given expansion:- ${(x-\frac{3}{x^2})}^9$

Here, $a=x,b=\frac{-3}{x^2},n=9$

Therefore,

$T_{r+1}={}^9{C}_r{x}^r{\left(\frac{-3}{x^2}\right)}^{9-r}$

$\Rightarrow T_{r+1}={}^9{C}_r{x}^{3(r-6)}{(-3)}^{9-r}$

For constant term,

$3(r-6)=0$

$\Rightarrow r=6$

Therefore,

Constant term $=T_7={}^9{C}_6{x}^0{(-3)}^{9-6}=-{}^9{C}_6{3}^3$