Question

The coefficient of $x^3$ in the expansion of $\left(x -\frac{1}{x}\right)^{7}$ is :

• A. 14
• B. 21
• C. 28
• D. 35

Answer 1

Answer: (B)

21

Answer 2

Given, $\left(x -\frac{1}{x}\right)^{7}$ and the $\left(r+1\right)^{th}$ term in the
expansion of $\left(x+a\right)^{n} T_{\left(r+1\right)} = \,^{n}C_{r} \left(x\right)^{n-r} a^{r}$
$\therefore\left(r+1\right)^{th}$ term in expansion of
$\left(x- \frac{1}{x} \right)^{7} = \,^{7}C_{r} \left(x\right)^{7-r} \left(- \frac{1}{x}\right)^{r}$
$=\,^{7}C_{r} \left(x\right)^{7-2r}\left(-1\right)^{r}$
Since $x^{3}$ occurs in $T_{r+1}$
$\therefore 7-2r =3 \Rightarrow r=2$
thus the coefficient of $x^{3} =\,^{7}C_{2} \left(-1\right)^{2}$
$= \frac{7\times6}{2\times1} =21$