Question

Show that the middle term in the expansion of ${(1+x)}^{2n}$ is
$\frac{\mathrm{1.3.5....}(2n-1)}{|\underset{–}{n}}{2}^n{x}^n$.

Answer 1

Number of terms $=2n$ which is even

So, middle term $=(n+1{)}^{th}$ term

Hence, we need $T_{n+1}$

General term of $(a+b{)}^n$ is

$T_{r+1}={}_r^{n}C{a}^{n-r}{b}^r$

For $T_{n+1},$ putting $n=2r,r=n,a=1,b=x$

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

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

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

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

Simplifying the expression, the middle term

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