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