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Question Number 43766 by peter frank last updated on 15/Sep/18

$$\mathrm{If}\mid a\mid <1\mathrm{and}\mid b\mid <1,\mathrm{then}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}$$ $$\mathrm{series}a(a+b)+a^2(a^2+b^2)+a^3(a^3+b^3)+...$$ $$\mathrm{upto}\mathrm{\infty }\mathrm{is}$$

Answered by MrW3 last updated on 15/Sep/18

$$a(a+b)+a^2(a^2+b^2)+a^3(a^3+b^3)+...$$ $$=(a^2+a^4+a^6+...)+[{\left(ab\right)}^2+{\left(ab\right)}^4+{\left(ab\right)}^6+...]$$ $$\to \frac{a^2}{1-a^2}+\frac{{\left(ab\right)}^2}{1-{\left(ab\right)}^2}$$ $$=\frac{a^2(1-a^4{b}^2)}{(1-a^2)(1-a^2{b}^2)}$$

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