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)

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 $