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Question Number 155242 by aliyn last updated on 27/Sep/21

the value of š›ƒ (2, n ) is ?

$$\boldsymbol{{the}}\:\boldsymbol{{value}}\:\boldsymbol{{of}}\:\:\boldsymbol{\beta}\:\left(\mathrm{2},\:{n}\:\right)\:\boldsymbol{{is}}\:? \\ $$

Commented by tabata last updated on 27/Sep/21

= ((Ī“2 Ī“(n))/(Ī“(n+2))) = (((nāˆ’1)!)/((n+1)!)) = (((nāˆ’1)!)/(n (n+1)(nāˆ’1)!)) = (1/(n (n+1)))

$$=\:\frac{\Gamma\mathrm{2}\:\Gamma\left({n}\right)}{\Gamma\left({n}+\mathrm{2}\right)}\:=\:\frac{\left({n}āˆ’\mathrm{1}\right)!}{\left({n}+\mathrm{1}\right)!}\:=\:\frac{\left({n}āˆ’\mathrm{1}\right)!}{{n}\:\left({n}+\mathrm{1}\right)\left({n}āˆ’\mathrm{1}\right)!}\:=\:\frac{\mathrm{1}}{{n}\:\left({n}+\mathrm{1}\right)} \\ $$

Answered by puissant last updated on 27/Sep/21

Ī²(2,n)=((Ī“(2)Ī“(n))/(Ī“(n+2)))=((1(nāˆ’1)!)/((n+1)!))=(1/(n(n+1))) .. =(1/n)āˆ’(1/(n+1))..

$$\beta\left(\mathrm{2},{n}\right)=\frac{\Gamma\left(\mathrm{2}\right)\Gamma\left({n}\right)}{\Gamma\left({n}+\mathrm{2}\right)}=\frac{\mathrm{1}\left({n}āˆ’\mathrm{1}\right)!}{\left({n}+\mathrm{1}\right)!}=\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)}\:.. \\ $$$$=\frac{\mathrm{1}}{{n}}āˆ’\frac{\mathrm{1}}{{n}+\mathrm{1}}.. \\ $$

Commented by aliyn last updated on 27/Sep/21

sir Ī“2 = Ī“(1+1) = 1 Ɨ 1! = 1 how Ī“2 = 2 ?

$${sir}\:\Gamma\mathrm{2}\:=\:\Gamma\left(\mathrm{1}+\mathrm{1}\right)\:=\:\mathrm{1}\:Ɨ\:\mathrm{1}!\:=\:\mathrm{1}\:{how}\:\Gamma\mathrm{2}\:=\:\mathrm{2}\:? \\ $$

Commented by puissant last updated on 27/Sep/21

Thanks..

$${Thanks}.. \\ $$

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