Question

Prove that: $\frac{(2n+1)!}{n!}=2^n\{1\cdot 3\cdot \mathrm{5...}(2n-1\left)\right(2n+1\left)\right\}$.

Answer 1

$\frac{(2n+1)!}{n!}=\frac{(2n+1)\left(2n\right)(2n-1)(2n-2)----\mathrm{5.4.3.2.1}}{\left(n\right)(n-1)---\mathrm{4.3.2.1}}$

$=\frac{((2n+1)(2n-1)---\mathrm{5.3.1})(\left(2n\right)(2n-2)---\mathrm{6.4.2})}{\left(n\right)(n-1)---\mathrm{4.3.2.1}}$

$=\frac{2^n((2n+1)(2n-1)---\mathrm{5.2.1})(\left(n\right)(n-1)---\mathrm{3.2.1})}{\left(n\right)(n-1)---\mathrm{4.3.2.1}}$

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