Question

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

Answer 1

LHS$=\frac{(2n+1)!}{n!}$

$=\frac{(2n+1)\left(2n\right)(2n-1)(2n-2)(2n-3)\dots \dots .6.5.4.3.2.1}{n!}$

$=\frac{2^n\left[n\right(n-1\left)\right(n-2)\dots ...3.2.1](2n-1)(2n-1)(2n-3)\dots \dots ..5.3.1}{n!}$

$=2^n.1.3.5.7...\dots (2n-1)(2n+1)$

$=$RHS

$\because n!=n(n-1)(n-2)(n-3)\dots \dots \mathrm{4.3.2.1}$.