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Question Number 192791 by Mastermind last updated on 27/May/23

$$\mathrm{If}\mathrm{n}\mathrm{is}\mathrm{a}\mathrm{positive}\mathrm{integer},\mathrm{prove}\mathrm{that}$$$$2^{\mathrm{n}}\mathrm{\Gamma }(\mathrm{n}+\frac{1}{2})=1.3.5...(2\mathrm{n}-1)\sqrt{\pi }.$$$$$$$$\mathrm{Help}!$$

Answered by Mathspace last updated on 27/May/23

$$montronsque\mathrm{\Gamma }(n+\frac{1}{2})=\frac{1.3.5...(2n-1)}{2^n}\sqrt{\pi }$$$$parrecurencesurn$$$$n=1\mathrm{\Gamma }\left(\frac{3}{2}\right)=\frac{\sqrt{\pi }}{2}$$$$vraiecar$$$$\mathrm{\Gamma }\left(\frac{3}{2}\right)=\mathrm{\Gamma }(1+\frac{1}{2})=\frac{1}{2}\mathrm{\Gamma }\left(\frac{1}{2}\right)=\frac{\sqrt{\pi }}{2}$$$$supposonslarelationvraiepourn$$$$ona\mathrm{\Gamma }(n+1+\frac{1}{2})$$$$=\mathrm{\Gamma }(1+n+\frac{1}{2})=(n+\frac{1}{2})\mathrm{\Gamma }(n+\frac{1}{2})$$$$=(n+\frac{1}{2})\times \frac{1.3.5...(2n-1)}{2^n}\sqrt{\pi }$$$$=\frac{1.3.5....(2n-1)(2n+1)}{2^{n+1}}\sqrt{\pi }$$$$larelationestvraiealordre$$$$n+1.$$

Commented by Mastermind last updated on 28/May/23

$$\mathrm{you}\mathrm{mean}\mathrm{it}\mathrm{should}\mathrm{be}(2\mathrm{n}+1)\mathrm{not}(2\mathrm{n}-1)?$$$$\mathrm{besides}\mathrm{pls}\mathrm{use}\mathrm{english}$$$$$$

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