Question

The sum of the coefficient of first 3 terms in the expansion ${\left(x-\frac{3}{x^2}\right)}^m$ in 559 . Find the term of the expansion containing $x^3$ .

Answer 1

Solution :-

Sum of coefficients of the first three terms of $(x-\frac{3}{x^2}{)}^m$

Now we know (- in the expansion $(a+b{)}^n$,

the general term is $T_{r+1}{=}^n{C}_r{a}^{n-r}{b}^r$

In expansion $(x-\frac{3}{x^2}{)}^m$, general term is $T_{r+1}{=}^m{C}_r{x}^{m-r}(\frac{-3}{x^2}{)}^r$

Now, first 3 terms are $T_1,T_2$ and $T_3$

$T_1{=}^m{C}_0{x}^m(\frac{-3}{x^2}{)}^0=x^m$

$T_2{=}^m{C}_1{x}^{m-1}(\frac{-3}{x^2}{)}^1=-3^m{C}_1{x}^{m-3}=-3^3{C}_1$

$T_3{=}^m{C}_2\left(x{)}^{m-2}\right(\frac{-3}{x^2}{)}^2{=}^m{C}_2{x}^{m-2}\left(\frac{(-3{)}^2}{x^4}\right)=9^m{C}_2.x^{m-6}$

$=9^m{C}_2$

Now given that, sum of first three = 559

$1-3^m{C}_1+9^m{C}_2=559$

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

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

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

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

$\Rightarrow (3m+31)(m-12)=0$

$\Rightarrow m=-\frac{31}{3}$ and $m=12$

$\because m$ is natural number $\therefore m=12$

Now, We need the coefficient of $x^3$

$T_{r+1}{=}^m{C}_r{x}^{m-r}\left(\frac{-3}{x^2}{)}^r{=}^{12}{C}_r{x}^{12-r}\right(\frac{-3}{x^2}{)}^r{=}^{12}{C}_r.(-3{)}^r.x^{12-3r}$

We need $x^3,$ $\therefore x^3=x^{12-3r}\Rightarrow r=3$

Hence, the term containing $x^3$

$T_{r+1}{=}^{12}{C}_r.(-3{)}^r.x^{12-3r}$

$T_{3+1}{=}^{12}{C}_3.(-3{)}^3{x}^{12-9}$

$=\frac{12!}{6!3!}\times -27\times x^3$

$T_4=-5940x^3$