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Question Number 61613 by naka3546 last updated on 05/Jun/19

$$S=1+\frac{3}{2018}+\frac{5}{{2018}^2}+\frac{7}{{2018}^3}+...$$$$4S-S^2=?$$

Commented by maxmathsup by imad last updated on 05/Jun/19

$$letS\left(x\right)=\sum _{n=}^{\mathrm{\infty }}{}_0(2n+1)x^n\Rightarrow S=S\left(\frac{1}{2018}\right)(\mid x\mid <1)$$$$S\left(x\right)=2\sum _{n=0}^{\mathrm{\infty }}{nx}^n+\sum _{n=0}^{\mathrm{\infty }}{x}^n$$$$\sum _{n=0}^{\mathrm{\infty }}{x}^n=\frac{1}{1-x}\Rightarrow \sum _{n=1}^{\mathrm{\infty }}{nx}^{n-1}=\frac{1}{{(1-x)}^2}\Rightarrow \sum _{n=1}^{\mathrm{\infty }}{nx}^n=\frac{x}{{(1-x)}^2}\Rightarrow$$$$S\left(x\right)=\frac{2x}{{(1-x)}^2}+\frac{1}{(1-x)}=\frac{2x+(1-x)}{{(1-x)}^2}=\frac{x+1}{{(1-x)}^2}\Rightarrow$$$$S=S\left(\frac{1}{2018}\right)=\frac{\frac{1}{2018}+1}{{(1-\frac{1}{2018})}^2}=\frac{2019}{2018{\left(\frac{2017}{2018}\right)}^2}=\frac{2018\times 2019}{{\left(2017\right)}^2}\Rightarrow$$$$4S-S^2=\frac{4\times 2018\times 2019}{{\left(2017\right)}^2}-\frac{{2018}^2\times {2019}^2}{{\left(2017\right)}^4}resttofinishthecalculus...$$$$$$

Answered by ajfour last updated on 05/Jun/19

$$S=1+\frac{3}{2018}+\frac{5}{{2018}^2}+\frac{7}{{2018}^3}+...$$$$2018S=2018+3+\frac{5}{2018}+\frac{7}{{2018}^2}+...$$$$\Rightarrow 2017S=2018+\frac{2}{1-\frac{1}{2018}}$$$$\Rightarrow S=\frac{2018}{2017}\times \frac{2019}{2017}.$$

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