Question

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

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

Answer 1

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

we have

$\begin{array}{c}{2}^n\left[\mathrm{1.3.5........}\left(2n-3\right)\left(2n-1\right)\right]\\ Multiplyanddivideby\mathrm{2.4.6....2}nweget\\ \left[\mathrm{1.3.5........}\left(2n-3\right)\left(2n-1\right)\right]2^n=\left[\mathrm{1.3.5........}\left(2n-3\right)\left(2n-1\right)\right]2^n\times \frac{\mathrm{2.4.6....2}n}{\mathrm{2.4.6....2}n}\\ =\frac{\left[\mathrm{1.2.3.4.5.......}\left(2n-1\right)\left(2n\right)\right]2^n}{\mathrm{2.4.6......2}n}\\ Onrearraningthenumerator,weget\\ =\frac{\left[\mathrm{1.2.3.4.5.......}\left(2n-1\right)\left(2n\right)\right]2^n}{\mathrm{2.4.6......2}n}\\ =\frac{\left[\left(2n\right)!\right]2^n}{\mathrm{2.4.6......2}n}\left\{\left(2n\right)!=\left(2n\right)\left(2n-1\right)..........5.4.3.2.1\right\}\\ Takingout2commonfromdenominator\\ =\frac{\left[\left(2!\right)\right]2^n}{\left[\mathrm{2.2.....}ntimes\right]\times \left[\mathrm{1.2.3.....}ntimes\right]}\\ =\frac{\left(2n\right)!2^n}{\left[2^n\right]\times \left[\left(n\right)\left(n+1\right)......3.2.1\right]}\\ =\frac{\left(2n\right)!}{n\left(n-1\right)......3.2.1}\\ =\frac{\left(2n\right)!}{n!}\\ Hence,theoptionAisthecorrectanswer.\end{array}$