Prove that the term independent of x in the expansion
Question:

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

Solution:

Given:

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

Suppose the term independent of $x$ is the $(r+1)$ th term.

$\therefore T_{r+1}={ }^{2 n} C_{r} x^{2 n-r} \frac{1}{x^{r}}$

$={ }^{2 n} C_{r} x^{2 n-2 r}$

For this term to be independent of $x$, we must have:

$2 n-2 r=0$

$\Rightarrow n=r$

$\therefore$ Required coefficient $={ }^{2 n} C_{n}$

$=\frac{(2 n) !}{(n !)^{2}}$

$=\frac{\{1 \cdot 3 \cdot 5 \ldots(2 n-3)(2 n-1)\}\{2 \cdot 4 \cdot 6 \ldots(2 n-2)(2 n)\}}{(n !)^{2}}$

$=\frac{\{1 \cdot 3 \cdot 5 \ldots(2 n-3)(2 n-1)\} 2^{n}}{n !}$