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., $-5940\,x^{3}$