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 $x^3={}^7{C}_2{\left(-1\right)}^2$
$=\frac{7\times 6}{2\times 1}=21$