Question

Prove that the term independent of x in the expansion of ${\left(x+\frac{1}{x}\right)}^{2n}$ is $\frac{1·3·5...\left(2n-1\right)}{n!}.2^n.$

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ {\left(x+\frac{1}{x}\right)}^{2n} \\ \mathrm{Suppose}\mathrm{the}\mathrm{term}\mathrm{independent}\mathrm{of}x\mathrm{is}\mathrm{the}(r+1)\mathrm{th}\mathrm{term}. \\ \therefore T_{r+1}={}^{2n}C_r{x}^{2n-r}\frac{1}{x^r} \\ ={}^{2n}C_r{x}^{2n-2r} \\ \mathrm{For}\mathrm{this}\mathrm{term}\mathrm{to}\mathrm{be}\mathrm{independent}\mathrm{of}x,\mathrm{we}\mathrm{must}\mathrm{have}: \\ 2n-2r=0 \\ \Rightarrow n=r \\ \therefore \mathrm{Required}\mathrm{coefficient}={}^{2n}C_n \\ =\frac{\left(2n\right)!}{(n!{)}^2} \\ =\frac{\left\{1·3·5...\left(2n-3\right)\left(2n-1\right)\right\}\left\{2·4·6...\left(2n-2\right)\left(2n\right)\right\}}{(n!{)}^2} \\ =\frac{\left\{1·3·5...\left(2n-3\right)\left(2n-1\right)\right\}{2}^n}{n!} \end{gathered}$$