Question

If $n$ is a positive integer, find the coefficient of $x^{-1}$ in the expansion of ${\left(1+x\right)}^n{\left(1+\frac{1}{x}\right)}^n$.

• A. $\frac{2n!}{\left(n-1\right)!\left(n+1\right)!}$
• B. $\frac{2n!}{{\left(n!\right)}^2}$
• C. $\frac{2n!}{n!\left(n-1\right)!}$
• D. $\frac{\left(2n-1\right)!}{n!\left(n-1\right)!}$

Answer 1

Answer: (A)

$\because {\left(1+x\right)}^n{\left(1+\frac{1}{x}\right)}^n$
$={\left(1+x\right)}^n{\left(\frac{x+1}{x}\right)}^n=\frac{{\left(1+x\right)}^{2n}}{x^n}$
Now to find the coefficient of $x^{-1}$ in ${\left(1+x\right)}^n{\left(1+\frac{1}{x}\right)}^n$, it is equivalent to find coefficient of $x^{-1}$ in $\frac{{\left(1+x\right)}^{2n}}{x^n}$ which in turn is equal to the coefficient of $x^{-1}$ in the expansion of ${\left(1+x\right)}^{2n}$
Since ${\left(1+x\right)}^2={}^{2n}{C}_0{x}^0+{}^{2n}{C}_1{x}^1+{}^{2n}{C}_2{x}^2+\dots$
$+{}^{2n}{C}_{n-1}{x}^{n-1}+\dots +{}^{2n}{C}_{2n}{x}^{2n}$
Thus, the coefficient of $x^{n-1}$ is ${}^{2n}{C}_{n-1}=\frac{\lfloor 2n}{\lfloor n-1\lfloor n+1}$