Question

The term independent of $x$ in the expansion of ${(x+\frac{1}{x})}^{2n}$ is

• A. $\frac{\mathrm{1.3.5...}(2n-1)}{n!}$
• B. $\frac{\mathrm{1.3.5...}(2n-1)}{n!}{2}^n$
• C. $\frac{\mathrm{1.3.5...}(2n+1)}{(n+1)!}$
• D. None of the above

Answer 1

The correct option is B $\frac{\mathrm{1.3.5...}(2n-1)}{n!}{2}^n$

Let $(r+1)th$ term is independent of $x$ in the expansion of ${(x+\frac{1}{x})}^{2n}$

$T_{r+1}{=}^{2n}{C}_r{x}^{2n-2r}$

This term is independent of $x$.

$\therefore 2n-2r=0$

$\Rightarrow r=n$

Thus, $T_{r+1}{=}^{2n}{C}_n$

$=\frac{\left(2n\right)!}{n!n!}=\frac{\mathrm{1.2.3..4}...(2n-1)\left(2n\right)}{n!n!}$

$=\frac{\left(\mathrm{1.3.5...}\right(2n-1\left)\right(\mathrm{1.2...}n)\cdot 2^n}{n!n!}$

$=\frac{\mathrm{1.3.5...}(2n-1)}{n!}\cdot 2^n$.