Question Number 127512 by mohammad17 last updated on 30/Dec/20
Commented by sts last updated on 30/Dec/20
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Answered by Dwaipayan Shikari last updated on 30/Dec/20
![β(n,n)=((Γ^2 (n))/(Γ(2n)))=((Γ^2 (n))/(Γ(n)Γ(n+(1/2))))2^(1−2n) (√π)=((Γ(n)Γ((1/2)))/(Γ(n+(1/2))))2^(1−2n) =β(n,(1/2))2^(1−2n) [AsΓ(n)Γ(n+(1/2))=2^(1−2n) (√π) Γ(2n)]](Q127518.png)
$$\beta\left({n},{n}\right)=\frac{\Gamma^{\mathrm{2}} \left({n}\right)}{\Gamma\left(\mathrm{2}{n}\right)}=\frac{\Gamma^{\mathrm{2}} \left({n}\right)}{\Gamma\left({n}\right)\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)}\mathrm{2}^{\mathrm{1}−\mathrm{2}{n}} \sqrt{\pi}=\frac{\Gamma\left({n}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)}\mathrm{2}^{\mathrm{1}−\mathrm{2}{n}} \\ $$$$=\beta\left({n},\frac{\mathrm{1}}{\mathrm{2}}\right)\mathrm{2}^{\mathrm{1}−\mathrm{2}{n}} \\ $$$$\left[{As}\Gamma\left({n}\right)\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{2}^{\mathrm{1}−\mathrm{2}{n}} \sqrt{\pi}\:\Gamma\left(\mathrm{2}{n}\right)\right] \\ $$