Question

Prove that : $\frac{\left(2n\right)!}{n!}=\left\{\mathrm{1.3.5....}(2n-1)\right\}{2}^n$.

Answer 1

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

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

Multiply and divide ${\mathrm{2.4.6.....2}}^n$ on R.H.S.

$\left[\mathrm{1.3.5.....}\right(2n-1\left)\right]2^n=\frac{\left[\mathrm{1.3.5.....}\right(2n-1\left)\right]2^n\times {\mathrm{2.4.6.....2}}^n}{{\mathrm{2.4.6......2}}^n}$

$=\frac{\mathrm{1.2.3.4.5...}\dots (2n-1)\left(2n^1\right)2^n}{{\mathrm{2.4.6.....2}}^n}$

On rearranging the numerator we get

$=\frac{\left[2n\right(2n-1)\dots ..5.4.3.2.1]2^n}{\mathrm{2.4.6.....2}n}$

$=\frac{\left(\right(2n)!)2^n}{{\mathrm{2.4.6.....2}}^n}$

Take out $2$ common from denominator

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

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