Question

In the expansion of ${\left[x-\frac{1}{x}\right]}^6$ the coefficient of $x^{\mathbf{0}}$ is

• A. $20$
• B. $-20$
• C. $30$
• D. $-30$

Answer 1

The correct option is B

$-20$

Determine the coefficient of $x^{\mathbf{0}}$ in the expansion of ${\left[x-\frac{1}{x}\right]}^6$

We know that the $r^{th}$ term in the binomial expansion of ${\left(a+b\right)}^n$ is $T_{r+1}={}^n{C}_r{a}^{n-r}{b}^r$

$$\begin{gathered} \therefore T_{r+1}={}^6{C}_r{\left(x\right)}^{6-r}{\left(-\frac{1}{x}\right)}^r \\ ={}^6{C}_r{\left(x\right)}^{6-r-r}{\left(-1\right)}^r \\ ={}^6{C}_r{\left(x\right)}^{6-2r}{\left(-1\right)}^r....\left(i\right) \end{gathered}$$

As we have to find the coefficient of $x^{\mathbf{0}}$ then

$$\begin{gathered} \Rightarrow 6-2r=0 \\ \Rightarrow r=3 \\ \Rightarrow {}^6{C}_r{\left(x\right)}^{6-2r}{\left(-1\right)}^r={}^6{C}_3{\left(x\right)}^{6-2\left(3\right)}{\left(-1\right)}^3[\mathrm{Substituting}r=3\mathrm{into}(i)] \\ ={-}^6{C}_3{x}^0 \\ =-\frac{6!}{3!(6-3)!}{x}^0 \\ =-20x^0 \end{gathered}$$

Hence, option B is correct .