Question Number 92777 by ckkim89 last updated on 09/May/20
$$\mathrm{a}_{\mathrm{n}+\mathrm{1}} =\left(\mathrm{2n}+\mathrm{1}\right)\mathrm{a}_{\mathrm{n}} \\ $$$$\mathrm{a}_{\mathrm{1}} =\mathrm{1} \\ $$$$\mathrm{a}_{\mathrm{n}} =? \\ $$$$ \\ $$
Commented by mr W last updated on 09/May/20
$${a}_{{n}} =\left(\mathrm{2}{n}−\mathrm{1}\right){a}_{{n}−\mathrm{1}} =\left(\mathrm{2}{n}−\mathrm{1}\right)...\mathrm{3}×\mathrm{1}=\left(\mathrm{2}{n}−\mathrm{1}\right)!! \\ $$
Commented by mathmax by abdo last updated on 09/May/20
$$\frac{{a}_{{n}+\mathrm{1}} }{{a}_{{n}} }\:=\left(\mathrm{2}{n}+\mathrm{1}\right)\:\Rightarrow\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\frac{{a}_{{k}+\mathrm{1}} }{{a}_{{k}} }\:=\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \left(\mathrm{2}{k}+\mathrm{1}\right)\:\Rightarrow \\ $$$$\frac{{a}_{\mathrm{2}} }{{a}_{\mathrm{1}} }×\frac{{a}_{\mathrm{3}} }{{a}_{\mathrm{2}} }×....×\frac{{a}_{{n}} }{{a}_{{n}−\mathrm{1}} }\:=\mathrm{3}×\mathrm{5}×\mathrm{7}×....\left(\mathrm{2}{n}−\mathrm{1}\right)\:\Rightarrow \\ $$$${a}_{{n}} =\mathrm{2}×\mathrm{3}×\mathrm{4}×\mathrm{5}×\mathrm{6}......\left(\mathrm{2}{n}−\mathrm{1}\right)×\mathrm{2}{n}×\frac{\mathrm{1}}{\mathrm{2}×\mathrm{4}×\mathrm{6}×....\left(\mathrm{2}{n}\right)} \\ $$$$=\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{{n}} ×{n}!}\:\Rightarrow\:{a}_{{n}} =\frac{\left(\mathrm{2}{n}\right)!}{{n}!\:×\mathrm{2}^{{n}} } \\ $$
Answered by prakash jain last updated on 09/May/20
$${a}_{\mathrm{1}} =\mathrm{1} \\ $$$${a}_{\mathrm{2}} =\mathrm{3} \\ $$$$... \\ $$$${a}_{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\mathrm{2}{k}+\mathrm{1}\right) \\ $$
Commented by ckkim89 last updated on 09/May/20
oh, thanks!:)