Question

The coefficient of $\frac{1}{x}$ in the expansion of ${\left(1+x\right)}^n{\left(1+\frac{1}{x}\right)}^n$ is
(a) $\frac{n!}{\left[\left(n-1\right)!\left(n+1\right)!\right]}$

(b) $\frac{\left(2n\right)!}{\left[\left(n-1\right)!\left(n+1\right)!\right]}$

(c) $\frac{\left(2n\right)!}{\left(2n-1\right)!\left(2n+1\right)!}$

(d) none of these

Answer 1

(b) $\frac{\left(2n\right)!}{\left[\left(n-1\right)!\left(n+1\right)!\right]}$

$$\begin{gathered} \mathrm{Coefficient}\mathrm{of}\frac{1}{x}\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{expansion}=\mathrm{Coefficient}\mathrm{of}1\mathrm{in}(1+x{)}^n\times \mathrm{Coefficient}\mathrm{of}\frac{1}{x}\mathrm{in}{\left(1+\frac{1}{x}\right)}^n+\mathrm{Coefficient}\mathrm{of}x\mathrm{in}(1+x{)}^n\times \mathrm{Coefficient}\mathrm{of}\frac{1}{x^2}\mathrm{in}{\left(1+\frac{1}{x}\right)}^n \\ ={}^{n}C_0\times {}^{n}C_1+{}^{n}C_1\times {}^{n}C_2 \\ =n+n\times \frac{n!}{2\left(n-2\right)!} \\ =n+n\frac{n\left(n-1\right)}{2} \\ = \end{gathered}$$