Question

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

• A. -5940
• B. 1010
• C. 1001
• D. 1002

Answer 1

The correct option is A -5940

$\begin{array}{c}{\left(x-\frac{3}{x^2}\right)}^m{=}^m{C}_o{\left(x\right)}^m{\left(\frac{-3}{x^2}\right)}^0{+}^m{C}_1{\left(x\right)}^{m-1}{\left(\frac{-3}{x^2}\right)}^1{+}^m{C}_2{x}^{m-2}{\left(\frac{-3}{x^2}\right)}^2+......\\ {=}^m{C}_o{\left(x\right)}^m+{\left(-3\right)}^m{C}_1{\left(x\right)}^{m-1-2}+{9.}^m{C}_2{\left(x\right)}^{m-2-4}+....\\ {\Rightarrow }^m{C}_o+{\left(-3\right)}^m{C}_1+{9.}^m{C}_2=559\\ \Rightarrow 1-3m+9\frac{m\left(m-1\right)}{2}=559\\ \Rightarrow 2-6m+9m^2-9m=2\times 559=1118\\ \Rightarrow 9m^2-15m-1116=0\\ \Rightarrow 3m^2-15m-372=0\\ \Rightarrow 3m^2-36m+31m-372=0\\ \Rightarrow 3m\left(m-12\right)+31\left(m-12\right)=0\\ \Rightarrow \left(m-12\right)\left(3m+31\right)=0\\ \therefore m=12,andm=\frac{-31}{3}\left(doesnotexist\right)\\ so,wetakem=12\\ {\left(x-\frac{3}{x^2}\right)}^{12}\\ Then\\ T_{r+1}{=}^{12}{C}_r{x}^{12r}{\left(\frac{-3}{x^2}\right)}^r\\ ={\left(-1\right)}^r{3^r}^{12}{C}_r{x}^{12-r-25}\\ ={\left(-1\right)}^r{3^r}^{12}{C}_r{x}^{12-3r}\\ Containg{x}^3\\ 12-3r=0\\ \therefore r=3\\ Now,\\ T_4=T_{3+1}={\left(-1\right)}^3{3}^3{}^{12}{C}_3{x}^3\\ =-27{}^{12}{C}_3{x}^3\\ T^4=-5940x^3\end{array}$

hence, the option $A$ is the correct answer.