Question

The value of $2^n\{1\times 3\times \mathrm{5.....}(2n-3)\times (2n-1\left)\right\}$ is $\frac{\left(2n\right)!}{n!}$

• A.
• B.

Answer 1

The correct option is A

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

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