Question

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

• A. $\frac{\left(2n\right)!}{n!}$
• B. $\frac{\left(2n\right)!}{2^n}$
• C. $\frac{n!}{\left(2n\right)!}$
• D. None of these

Answer 1

The correct option is A

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

Solve the series:

The given expression is

$P=2^n\left\{1.3.5......\left(2n-3\right)\left(2n-1\right)\right\}$

Multiply and divide by $\left(2.4.6......2n\right)$

$P=\frac{2^n\left\{1.3.5......\left(2n-3\right)\left(2n-1\right)\right\}(2.4.6.....2n)}{2.4.6......2n}$

$=\frac{2^n\left\{1.2.3.4.5.6.......\left(2n-3\right)\left(2n-2\right)\left(2n-1\right)\left(2n\right)\right\}}{(2\times 2\times 2....ntimes)(1\times 2\times 3.....\times n)}$

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

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

Hence, the value of $2^n\left\{1.3.5......\left(2n-3\right)\left(2n-1\right)\right\}$ is $\frac{\left(2n\right)!}{n!}$.

Hence, option (A) is the correct answer.