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Question Number 83543 by Power last updated on 03/Mar/20

Commented by john santu last updated on 03/Mar/20

$$\mathrm{let}1+2\mathrm{x}+{3\mathrm{x}}^2+{4\mathrm{x}}^3+{5\mathrm{x}}^4+...=\mathrm{f}\left(\mathrm{x}\right)$$$$\int \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\int (1+2\mathrm{x}+{3\mathrm{x}}^2+{4\mathrm{x}}^3+...)\mathrm{dx}$$$$\int \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\mathrm{x}+{\mathrm{x}}^2+{\mathrm{x}}^3+{\mathrm{x}}^4+...$$$$\int \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\frac{\mathrm{x}}{1-\mathrm{x}}=\frac{\mathrm{x}}{-\mathrm{x}+1}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{1}{{(1-\mathrm{x})}^2}$$$$\frac{1}{{(1-\mathrm{x})}^2}=\frac{49}{36}\Rightarrow {(1-\mathrm{x})}^2=\frac{36}{49}$$$$(1-\mathrm{x}-\frac{6}{7})(1-\mathrm{x}+\frac{6}{7})=0$$$$(\frac{1}{7}-\mathrm{x})(\frac{13}{7}-\mathrm{x})=0$$$$\mathrm{x}=\{\begin{array}{l}\frac{1}{7}\\ \frac{13}{7}\end{array}$$

Commented by Power last updated on 03/Mar/20

$$\mathrm{thanks}$$

Answered by mr W last updated on 03/Mar/20

$$f\left(x\right)=1+2x+3x^2+4x^3+5x^4+...$$$$xf\left(x\right)=x+2x^2+3x^3+4x^4+...$$$$(1-x)f\left(x\right)=1+x+x^2+x^3+x^4+...$$$$(1-x)f\left(x\right)=\frac{1}{1-x},if\mid x\mid <1$$$$(1-x)f\left(x\right)=\mathrm{\infty },if\mid x\mid ⩾1$$$$$$$$with\mid x\mid <1:$$$$\Rightarrow f\left(x\right)=\frac{1}{{(1-x)}^2}=\frac{49}{36}$$$$\Rightarrow \frac{1}{1-x}=\pm \frac{7}{6}$$$$\Rightarrow 1-x=\pm \frac{6}{7}$$$$\Rightarrow x=1\pm \frac{6}{7}$$$$since\mid x\mid <1,thereisonlyonesolution:$$$$\Rightarrow x=\frac{1}{7}$$

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