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Question Number 98271 by Ar Brandon last updated on 12/Jun/20

$$\mathcal{G}{\mathrm{ivenU}}_{\mathrm{n}}={\int }_0^1{\mathrm{x}}^{\mathrm{n}}\sqrt{1-\mathrm{x}}\mathrm{dx}\mathrm{n}\in \mathbb{N},\mathrm{show}\mathrm{that}$$$${\mathrm{U}}_{\mathrm{n}}=\frac{2^{\mathrm{n}+2}\mathrm{n}!(\mathrm{n}+1)}{(2\mathrm{n}+3)!}$$

Commented by Ar Brandon last updated on 12/Jun/20

Thanks �� But, it really isn't clear to me. Any explanations please ?��

Commented by PRITHWISH SEN 2 last updated on 12/Jun/20

$$\mathrm{then}{\mathrm{U}}_{\mathrm{n}}=\frac{2\mathrm{n}}{2\mathrm{n}+3}.\frac{2(\mathrm{n}-1)}{2\mathrm{n}+1}......\frac{4}{7}.\frac{2}{5}.\frac{2}{3}$$$$=\frac{2^{\mathrm{n}}\{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3)......2.1\}.2}{(2\mathrm{n}+3)(2\mathrm{n}+1).........7.5.3}$$$$=\frac{2^{\mathrm{n}+1}.\mathrm{n}!}{(2\mathrm{n}+3)!}\times 2^{\mathrm{n}+1}\{(\mathrm{n}+1)\mathrm{n}(\mathrm{n}-1)........2.1\}$$$$=\frac{2^{2\mathbf{n}+2}\mathbf{n}!(\mathbf{n}+1)!}{(2\mathbf{n}+3)!}\mathbf{I}\mathbf{got}\mathbf{this}\mathbf{please}\mathbf{check}$$

Commented by Ar Brandon last updated on 12/Jun/20

$${\mathrm{I}}^{\prime }\mathrm{m}\mathrm{stucked}\mathrm{here}.\mathrm{Can}\mathrm{anyone}\mathrm{help}\mathrm{me}\mathrm{out},\mathrm{please}?$$

Commented by Ar Brandon last updated on 12/Jun/20

Commented by PRITHWISH SEN 2 last updated on 20/Jul/20

$$\mathrm{I}\mathrm{did}\mathrm{not}\mathrm{do}\mathrm{anything}\mathrm{I}\mathrm{just}\mathrm{expanded}\mathrm{your}\mathrm{expression}$$$${\mathrm{U}}_{\mathrm{n}}=\frac{2\mathrm{n}}{2\mathrm{n}+3}{\mathrm{U}}_{\mathrm{n}-1}\mathrm{then}{\mathrm{U}}_{\mathrm{n}-1}=\frac{2(\mathrm{n}-1)}{2(\mathrm{n}-1)+3}{\mathrm{U}}_{\mathrm{n}-2}$$$$\Rightarrow {\mathrm{U}}_{\mathrm{n}-1}=\frac{2(\mathrm{n}-1)}{2\mathrm{n}+1}{\mathrm{U}}_{\mathrm{n}-2}\mathrm{and}\mathrm{likewise}\mathrm{you}\mathrm{can}\mathrm{get}$$$$\mathrm{by}\mathrm{putting}\mathrm{n}=\mathrm{n}-2,\mathrm{n}-3\mathrm{and}\mathrm{so}\mathrm{on}$$$$\mathrm{now}\mathrm{for}{\mathrm{U}}_1=\frac{2}{5}{\mathrm{U}}_0{\mathrm{andU}}_0={\int }_0^1{\mathrm{x}}^0\sqrt{1-\mathrm{x}}\mathrm{dx}=\frac{2}{3}$$$$\mathrm{I}\mathrm{think}\mathrm{now}\mathrm{it}\mathrm{makes}\mathrm{sense}$$

Answered by smridha last updated on 12/Jun/20

$$\mathit{l}\mathit{e}\mathit{t}\mathit{x}=\sin {}^2\mathit{\theta }\mathit{s}\mathit{o}$$$$2{\int }_0^{\frac{\mathit{\pi }}{2}}{\mathit{s}\mathit{i}\mathit{n}}^{(2\mathit{n}+1)}\mathit{\theta }{\mathit{c}\mathit{o}\mathit{s}}^2\mathit{\theta }\mathit{d}\mathit{\theta }$$$$=\frac{\mathbf{\Gamma }(\mathit{n}+1).\mathbf{\Gamma }\left(\frac{3}{2}\right)}{\mathbf{\Gamma }(\mathit{n}+\frac{5}{2})}=\frac{\mathit{n}!\sqrt{\mathit{\pi }}}{(2\mathit{n}+3).\mathbf{\Gamma }(\mathit{n}+\frac{3}{2})}$$$$\mathit{I}\mathit{t}\mathit{h}\mathit{i}\mathit{n}\mathit{k}\mathit{n}\mathit{o}\mathit{w}\mathit{y}\mathit{o}\mathit{u}\mathit{c}\mathit{a}\mathit{n}\mathit{e}\mathit{a}\mathit{s}\mathit{y}\mathit{l}\mathit{y}$$$$\mathit{g}\mathit{e}\mathit{t}\mathit{y}\mathit{o}\mathit{u}\mathit{r}\mathit{a}\mathit{n}\mathit{s}.$$

Commented by PRITHWISH SEN 2 last updated on 13/Jun/20

$$\mathrm{wow}!\mathrm{excellent}\mathrm{sir}.$$

Commented by smridha last updated on 13/Jun/20

$$\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{s}..$$

Commented by PRITHWISH SEN 2 last updated on 13/Jun/20

$$\mathrm{welcome}$$

Commented by Ar Brandon last updated on 13/Jun/20

Hum, thank you ��

Answered by Dwaipayan Shikari last updated on 01/Nov/20

$${\int }_0^1{x}^n{(1-x)}^{\frac{1}{2}}dx$$$$={\int }_0^1{x}^{n+1-1}{x}^{\frac{3}{2}-1}dx$$$$=\beta (n+1,\frac{3}{2})=\frac{\mathrm{\Gamma }(n+1)\mathrm{\Gamma }\left(\frac{3}{2}\right)}{\mathrm{\Gamma }(n+\frac{5}{2})}$$

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