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

Answer 1

The correct option is A

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

$\mathrm{1.3.5......}(2n-1)2^n$ = $\frac{\mathrm{1.2.3.4.5.6....}(2n-1)\left(2n\right)2^n}{\mathrm{2.4.5.....2}n}$

= $\frac{\left(2n\right)!2^n}{2^n\left(\mathrm{1.2.3......}n\right)}$ = $\frac{\left(2n\right)!}{n!}$