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 coefficient of the term in the expansion containing $x^3$.

• A. $-5940$
• B. $-5900$
• C. $-5945$
• D. $-5950$

Answer 1

Answer: (A)

The coefficients of the first three terms of ${\left(x-\frac{3}{x^2}\right)}^m$ are ${}^m{C}_0$, ${}^m{C}_1\left(-3\right)$ and ${}^m{C}_{2}9$.
Given, ${}^m{C}_0-3\times {}^m{C}_1+9\times {}^m{C}_2=559$
$\Rightarrow 1-3m+\frac{9m\left(m-1\right)}{2}=559$
which gives $m=12$ ($m$ being a natural number).
Now $T_{r+1}={}^{12}{C}_r{x}^{12-r}\left(-\frac{3}{x^2}\right)$
$={}^{12}{C}_r{\left(-3\right)}^r\cdot x^{12-3r}$
Put $12-3r=3$
$\Rightarrow r=3$.
Thus, the required term is ${}^{12}{C}_3{\left(-3\right)}^3{x}^3$, i.e., $-5940x^3$