Question

Show that the middle term in the expansion of $(1+x{)}^{2n}$ is $\frac{\mathrm{1.3.5.....}(2n-1)}{n!}$ $2^n{x}^n$; where n is a positive integer.

Answer 1

In the expansion of ${\left(1+x\right)}^{2n}$ in total 2n+1 terms are present.

So, middle term will be,

$\begin{array}{c}\frac{\left(2n+1\right)+1}{2}=\frac{2n+2}{2}\\ =n+1\end{array}$

So, middle term will be (n+1)th term.

So,

$\begin{array}{c}{a}_n{=}^{2n}{C}_n{\left(1\right)}^n{\left(x\right)}^{2n-n}\\ =\frac{2n!}{n!\left(2n-n\right)!}{x}^n\\ =\frac{2^n\left(n!\right)\mathrm{1.3.5....}\left(2n-1\right)}{n!n!}{x}^n\\ =\frac{\mathrm{1.3.5....}\left(2n-1\right)}{n!}{2}^n{x}^n\end{array}$

Hence proved.