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Question Number 76913 by Ajao yinka last updated on 31/Dec/19

Answered by ~blr237~ last updated on 01/Jan/20

$$\mathrm{let}\mathrm{it}\mathrm{be}\mathrm{A}$$$$1+\mathrm{x}+....+{\mathrm{x}}^{100}=\frac{1-{\mathrm{x}}^{101}}{1-\mathrm{x}}$$$$\mathrm{A}={\int }_0^1\frac{{\mathrm{x}}^{49}(1-\mathrm{x})}{1-{\mathrm{x}}^{101}}\mathrm{dx}={\int }_0^1{\mathrm{x}}^{49}(1-\mathrm{x})\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{\mathrm{x}}^{101\mathrm{n}}\mathrm{dx}$$$$\mathrm{A}=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{\int }_0^1{\mathrm{x}}^{101\mathrm{n}+49}(1-\mathrm{x})\mathrm{dx}$$$$=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\mathrm{B}(101\mathrm{n}+50,2)$$$$=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{\mathrm{\Gamma }(101\mathrm{n}+50)\mathrm{\Gamma }\left(2\right)}{\mathrm{\Gamma }(101\mathrm{n}+52)}$$$$=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{(101\mathrm{n}+51)(101\mathrm{n}+50)}=\frac{1}{{\left(101\right)}^2}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{(\mathrm{n}+\frac{51}{101})(\mathrm{n}+\frac{50}{101})}$$$$=\frac{1}{{101}^2}\times \frac{\psi \left(\frac{51}{101}\right)-\psi \left(\frac{50}{101}\right)}{\frac{51}{101}-\frac{50}{101}}=\frac{\psi \left(\frac{51}{101}\right)-\psi \left(\frac{50}{101}\right)}{101}\mathrm{with}\psi =\frac{{\mathrm{\Gamma }}^{\prime }}{\mathrm{\Gamma }}$$$$$$

Commented by jagoll last updated on 01/Jan/20

$$what\psi functionsir?$$

Commented by 20092104 last updated on 10/Feb/20

$$digammafunction$$$$\psi \left(x\right)=\frac{d}{dx}\ln \left(\mathrm{\Gamma }\left(x\right)\right)=\frac{{\mathrm{\Gamma }}^{\prime }\left(x\right)}{\mathrm{\Gamma }\left(x\right)}$$

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