Question

The coefficient of $\frac{1}{x}$ in the expansion of $(1+x{)}^n{(1+\frac{1}{x})}^n$ is

• A. $\frac{n!}{\left\{\right(n-1)!(n+1)!\}}$
• B. $\frac{\left(2n\right)!}{\left[\right(n-1)!(n+1)!]}$
• C. $\frac{\left(2n\right)!}{(2n-1)!(2n+1)!}$
• D. None of these

Answer 1

The correct option is B

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

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

$(1+x{)}^n{(1+\frac{1}{x})}^n=(1+x{)}^n$

${\left(\frac{x+1}{x}\right)}^n=\frac{(1+x{)}^{2n}}{x^n}$

Now to find the coefficient of $\frac{1}{x}in(1+x{)}^n{(1+\frac{1}{x})}^n$

$=\frac{1}{x^n}({}^{2n}{C}_0{x}^0+{}^{2n}{C}_1{x}^1+{}^{2n}{C}_2{x}^2+....+{}^{2n}{C}_{n-1}{x}^{n-1}+....+{}^{2n}{C}_{2n-1}{x}^{2n-1}+{}^{2n}{C}_{2n}{x}^{2n})$

$\therefore$ Coefficient of $\frac{1}{x}={}^{2n}{C}_{n-1}$

$[\because \frac{1}{x^n}\times x^{n-1}=\frac{1}{x}]$

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