Question Number 155242 by aliyn last updated on 27/Sep/21
$$\boldsymbol{{the}}\:\boldsymbol{{value}}\:\boldsymbol{{of}}\:\:\boldsymbol{\beta}\:\left(\mathrm{2},\:{n}\:\right)\:\boldsymbol{{is}}\:? \\ $$
Commented by tabata last updated on 27/Sep/21
$$=\:\frac{\Gamma\mathrm{2}\:\Gamma\left({n}\right)}{\Gamma\left({n}+\mathrm{2}\right)}\:=\:\frac{\left({n}ā\mathrm{1}\right)!}{\left({n}+\mathrm{1}\right)!}\:=\:\frac{\left({n}ā\mathrm{1}\right)!}{{n}\:\left({n}+\mathrm{1}\right)\left({n}ā\mathrm{1}\right)!}\:=\:\frac{\mathrm{1}}{{n}\:\left({n}+\mathrm{1}\right)} \\ $$
Answered by puissant last updated on 27/Sep/21
$$\beta\left(\mathrm{2},{n}\right)=\frac{\Gamma\left(\mathrm{2}\right)\Gamma\left({n}\right)}{\Gamma\left({n}+\mathrm{2}\right)}=\frac{\mathrm{1}\left({n}ā\mathrm{1}\right)!}{\left({n}+\mathrm{1}\right)!}=\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)}\:.. \\ $$$$=\frac{\mathrm{1}}{{n}}ā\frac{\mathrm{1}}{{n}+\mathrm{1}}.. \\ $$
Commented by aliyn last updated on 27/Sep/21
$${sir}\:\Gamma\mathrm{2}\:=\:\Gamma\left(\mathrm{1}+\mathrm{1}\right)\:=\:\mathrm{1}\:Ć\:\mathrm{1}!\:=\:\mathrm{1}\:{how}\:\Gamma\mathrm{2}\:=\:\mathrm{2}\:? \\ $$
Commented by puissant last updated on 27/Sep/21
$${Thanks}.. \\ $$