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Question Number 193435 by Mingma last updated on 14/Jun/23

Commented by maths_plus last updated on 14/Jun/23

$$\mathrm{i}\mathrm{want}\mathrm{to}\mathrm{known}...$$

Answered by qaz last updated on 14/Jun/23

$$\frac{1}{x}=y,y\in (-1,1)$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{ny}^n=\underset{n=0}{\sum ^{\mathrm{\infty }}}(n+1)y^{n+1}=y(yD+1)\frac{1}{1-y}=\frac{y}{{(1-y)}^2}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{n}^2{y}^n=\underset{n=0}{\sum ^{\mathrm{\infty }}}{(n+1)}^2{y}^{n+1}=y{(yD+1)}^2\frac{1}{1-y}=y\left(\frac{2y^2+3y(1-y)+{(1-y)}^2}{{(1-y)}^3}\right)$$$$\frac{y^2}{{(1-y)}^4}=\frac{y}{{(1-y)}^3}(2y^2+3y(1-y)+{(1-y)}^2)$$$$\Rightarrow y=\frac{\sqrt{5}-1}{2}\Rightarrow x=\frac{\sqrt{5}+1}{2}$$

Commented by Mingma last updated on 14/Jun/23

Perfect ��

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