Question Number 182584 by Matica last updated on 11/Dec/22
$$\:\:{Transform}\:{into}\:{factorial}\:{form}\:{the}\:{following} \\ $$$$\:\:{A}\:=\:\mathrm{1}×\mathrm{3}×\mathrm{5}×\mathrm{7}×\mathrm{9}×…×\left(\mathrm{2}{n}−\mathrm{1}\right)\:.\: \\ $$$${Example}:\:\mathrm{4}×\mathrm{3}×\mathrm{2}×\mathrm{1}=\mathrm{4}! \\ $$
Answered by aleks041103 last updated on 11/Dec/22
$${A}=\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}…\left(\mathrm{2}{n}−\mathrm{1}\right)=\frac{\mathrm{1}.\mathrm{2}.\mathrm{3}.\mathrm{4}.\mathrm{5}.\mathrm{6}.\mathrm{7}…\left(\mathrm{2}{n}−\mathrm{1}\right).\mathrm{2}{n}}{\mathrm{2}.\mathrm{4}.\mathrm{6}….\left(\mathrm{2}{n}\right)}= \\ $$$$=\frac{\left(\mathrm{2}{n}\right)!}{\left(\mathrm{2}.\mathrm{1}\right).\left(\mathrm{2}.\mathrm{2}\right).\left(\mathrm{2}.\mathrm{3}\right)….\left(\mathrm{2}.{n}\right)}=\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{{n}} {n}!} \\ $$$$\Rightarrow{A}=\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{{n}} {n}!} \\ $$