Question

The value of $2^n\left[\mathrm{1.3.5...}\right(2n-3\left)\right(2n-1\left)\right]$ is

• A. $\frac{\left(2n\right)!}{n!}$
• B. $\frac{\left(2n\right)!}{2^n}$
• C. $\frac{n!}{\left(2n\right)!}$
• D. $Noneofthese$

Answer 1

The correct option is A $\frac{\left(2n\right)!}{n!}$

$2^n\left[\mathrm{1.3.5....}\right(2n-3\left)\right(2n-1\left)\right]$

$\mathrm{1.3.5.7.....}(2n-1)=\frac{\left(1\right)\left(2\right)\left(3\right)\left(4\right)\left(5\right)\dots \dots \left(2n\right)}{\left(2\right)\left(4\right)\left(6\right)\dots ..\left(2n\right)}$

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

$\mathrm{2.4.6.8}=2\times 2\times 2\times (3\times 2)\times (4\times 2)$

$=2^n\times n!$

$\therefore 1\times 3\times 5\times \mathrm{7....}(2n-1)=\frac{2n!}{2^n\times n!}$

$\therefore 2^n\left[\mathrm{1.3.5.....}\right(2n-3\left)\right(2n-1\left)\right]$

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