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Question Number 67826 by peter frank last updated on 31/Aug/19

Commented by gunawan last updated on 01/Sep/19

$$x=1+a+a^2+a^3+..$$$$x=\frac{1}{1-a}\Rightarrow 1-a=\frac{1}{x}\Rightarrow a=1-\frac{1}{x}=\frac{x-1}{x}$$$$y=1+b+b^2+b^3+...$$$$y=\frac{1}{1-b}\Rightarrow 1-b=\frac{1}{y}\Rightarrow b=\frac{y-1}{y}$$$$1+\left(ab\right)+{\left(ab\right)}^2+{\left(ab\right)}^3+..$$$$=\frac{1}{1-ab}=\frac{1}{\frac{xy}{xy}-\left(\frac{(x-1)(y-1)}{xy}\right)}$$$$=\frac{1}{\left(\frac{xy-(xy-x-y+1)}{xy}\right)}=\frac{xy}{x+y-1}$$

Commented by mathmax by abdo last updated on 01/Sep/19

$$itsnotcorrectif\mid a\mid ⩾1or\mid b\mid ⩾1...!$$

Commented by peter frank last updated on 01/Sep/19

$$thankyou$$

Commented by peter frank last updated on 01/Sep/19

$$thankyou$$

Answered by mind is power last updated on 01/Sep/19

$$x=\frac{1}{1-a}\Rightarrow a=\frac{x-1}{x}$$$$y=\frac{1}{1-b}\Rightarrow b=\frac{y-1}{y}$$$$\Rightarrow 1+ab+a^2{b}^2+.....=\mathrm{\Sigma }{\left(ab\right)}^k={\left(\left(\frac{y-1}{y}\right)\left(\frac{x-1}{x}\right)\right)}^k=\frac{1}{1-\left(\frac{x-1}{x}\right)\left(\frac{y-1}{y}\right)}=\frac{xy}{xy-xy+y+x-1}=\frac{xy}{x+y-1}$$

Answered by rrebo5637@gmail.com last updated on 01/Sep/19

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