Question

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

Answer 1

We have,
LHS = $\frac{(2n+1)!}{n!}$
= $\frac{(2n+1)\left[\mathrm{1.2.3.4.5.6.7.8...}\right(2n-1\left)2n\right]}{n!}=\frac{\left[\mathrm{1.3.5.7...}\right(2n-1\left)\right(2n+1\left)\right]2"\times n!}{n!}$
=$2^n\left[\mathrm{1.3.5.7...}\right(2n-1\left)\right(2n+1\left)\right]$
=RHS Hence proved.