Question

The sum of the coefficients of first three terms in the expansion of ${(x-\frac{3}{x^2})}^m,x\ne 0$, $m$ being a natural number, is $559$. Find the term of the expansion containing $x^3$.

Answer 1

Given, ${(x-\frac{3}{x^2})}^m$

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

$T_{r+1}{=}^m{C}_r(-3{)}^r.x^{m-3r}$

Given,

${}^m{c}_0(-3{)}^0{+}^m{C}_1(-3{)}^1{+}^m{C}_2(-3{)}^2=559$

$\Rightarrow 1+m(-3)+9\times \frac{m(m-1)}{2}=559$

$\Rightarrow 2-6m+9m^2-9m=559\times 2$

$\Rightarrow 9m^2-15m-1116=0$

$m=12,m=\frac{-31}{3}\left(Impossible\right)$

$\therefore m=12$

$Tr+1=12C_r(-3{)}^r{x}^{12-3r}$

$12-3r=3$

$\therefore r=3$

$T_3=12C_2(-3{)}^2$

$T_3=594$