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Question Number 143846 by mathmax by abdo last updated on 18/Jun/21

$$\mathrm{find}\mathrm{the}\mathrm{sum}\frac{{(0!)}^2}{1!}+\frac{{(1!)}^2}{3!}+\frac{{(2!)}^2}{5!}+.....$$

Answered by Ar Brandon last updated on 18/Jun/21

$$\frac{(\mathrm{n}!)(\mathrm{n}!)}{(\mathrm{n}+\mathrm{n}+1)!}=\frac{\mathrm{\Gamma }(\mathrm{n}+1)\mathrm{\Gamma }(\mathrm{n}+1)}{\mathrm{\Gamma }(2\mathrm{n}+2)}=\beta (\mathrm{n}+1,\mathrm{n}+1)$$$$\mathrm{S}=\frac{{(0!)}^2}{1!}+\frac{{(1!)}^2}{3!}+\frac{{(2!)}^2}{5!}+\cdot \cdot \cdot =\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\beta (\mathrm{n}+1,\mathrm{n}+1)$$$$={\int }_0^1\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{\mathrm{x}}^{\mathrm{n}}{(1-\mathrm{x})}^{\mathrm{n}}={\int }_0^1\frac{1}{1-(\mathrm{x}-{\mathrm{x}}^2)}\mathrm{dx}$$$$={\int }_0^1\frac{\mathrm{dx}}{{\mathrm{x}}^2-\mathrm{x}+1}={\int }_0^1\frac{\mathrm{dx}}{{(\mathrm{x}-\frac{1}{2})}^2+\frac{3}{4}}$$$$={\left[\frac{2}{\sqrt{3}}\arctan \left(\frac{2\mathrm{x}-1}{\sqrt{3}}\right)\right]}_0^1=\frac{2}{\sqrt{3}}(\frac{\pi }{6}+\frac{\pi }{6})=\frac{2\sqrt{3}\pi }{9}$$

Commented by mathmax by abdo last updated on 19/Jun/21

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Commented by Ar Brandon last updated on 19/Jun/21

$$\mathrm{Your}\mathrm{welcome},\mathrm{Sir}.$$

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