Question

The middle term in the expansion of ${(1+x)}^{2n}$ is , $n$ being a positive integer is

• A. $\frac{\left\{\mathrm{1.3.5....}\right(2n\left)\right\}{2}^n}{n!}{x}^n$
• B. $\frac{\left\{\mathrm{1.3.5....}\right(2n\left)\right\}{2}^{n}n!}{n!}{x}^n$
• C. $\frac{\left\{\mathrm{1.3.5....}\right(2n-1\left)\right\}{2}^{n}n!}{n!}{x}^n$
• D. $\frac{\left\{\mathrm{1.3.5....}\right(2n-1\left)\right\}{2}^n}{n!}{x}^n$

Answer 1

The correct option is D $\frac{\left\{\mathrm{1.3.5....}\right(2n-1\left)\right\}{2}^n}{n!}{x}^n$

The number of terms in the expansion of ${(1+x)}^{2n}$ is $(2n+1)$ (odd), its middle term is $\frac{2n+1+1}{2}=(n+1{)}^{th}$ term.$\therefore$ Required term $=T_{n+1}={{}^{2n}C}_n{x}^n=\frac{2n!}{n!n!}{x}^n=\frac{\left\{\mathrm{1.2.3.4.5.6...}\right(2n-1\left)2n\right\}}{n!n!}{x}^n=\frac{\left\{\mathrm{1.3.5....}\right(2n-1\left)\right\}\left\{\mathrm{2.4.6.....2}\right\}n}{n!n!}{x}^n=\frac{\left\{\mathrm{1.3.5....}\right(2n-1\left)\right\}{2}^n\left(\mathrm{1.2.3....}n\right)}{n!n!}{x}^n=\frac{\left\{\mathrm{1.3.5....}\right(2n-1\left)\right\}{2}^n}{n!}{x}^n$