Question

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

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

Answer 1

We have,

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

$\therefore T_{r+1}={}^{2n}{C}_r(x{)}^{2n-r}{\left(\frac{1}{x}\right)}^r$

$={}^{2n}{C}_r(x{)}^{2n-r-r}$

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

If it is independent of x, we must have,

2n-2r=0

$\Rightarrow 2n=2r$

$\Rightarrow f=n$

$\therefore$ Term independent of $x=T_{n+1}$

Now,

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

$={}^{2n}{C}_n$

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

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

$=\frac{\left(2n\right)(2n-1)(2n-2)...5\times 4\times 3\times 2\times 1}{n!n!}$

$=\frac{\{1\times 3\times 5\times ....(2n-1\left)\right\}\{2\times 4\times 6\times ....2n\}}{n!n!}$

$=\frac{\{1\times 3\times 5\times ...(2n-1\left)\right\}\times 2^n\{1\times 2\times 3\times ...n\}}{n!n!}$

$=\frac{\{1\times 3\times 5\times ....(2n-1\left)\right\}\times 2^n\times n!}{n!n!}$

$=2^n\times \frac{\{1\times 3\times 5\times ....(2n-1\left)\right\}}{n!}$

$\therefore$ The term independent to x

$=\frac{\{1\times 3\times 5\times ...(2n-1\left)\right\}}{n!}\times 2^n$

Hence proved.