Question Number 43766 by peter frank last updated on 15/Sep/18
$$\mathrm{If}\:\mid{a}\mid<\mathrm{1}\:\mathrm{and}\:\mid{b}\mid<\mathrm{1},\:\mathrm{then}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{series}\:{a}\left({a}+{b}\right)+{a}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)+{a}^{\mathrm{3}} \left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} \right)+… \\ $$$$\mathrm{upto}\:\infty\:\mathrm{is} \\ $$
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 +...)+[(ab)^2 +(ab)^4 +(ab)^6 +...] →(a^2 /(1−a^2 ))+(((ab)^2 )/(1−(ab)^2 )) =((a^2 (1−a^4 b^2 ))/((1−a^2 )(1−a^2 b^2 )))](https://www.tinkutara.com/question/Q43772.png)
$${a}\left({a}+{b}\right)+{a}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)+{a}^{\mathrm{3}} \left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} \right)+… \\ $$$$=\left({a}^{\mathrm{2}} +{a}^{\mathrm{4}} +{a}^{\mathrm{6}} +…\right)+\left[\left({ab}\right)^{\mathrm{2}} +\left({ab}\right)^{\mathrm{4}} +\left({ab}\right)^{\mathrm{6}} +…\right] \\ $$$$\rightarrow\frac{{a}^{\mathrm{2}} }{\mathrm{1}−{a}^{\mathrm{2}} }+\frac{\left({ab}\right)^{\mathrm{2}} }{\mathrm{1}−\left({ab}\right)^{\mathrm{2}} } \\ $$$$=\frac{{a}^{\mathrm{2}} \left(\mathrm{1}−{a}^{\mathrm{4}} {b}^{\mathrm{2}} \right)}{\left(\mathrm{1}−{a}^{\mathrm{2}} \right)\left(\mathrm{1}−{a}^{\mathrm{2}} {b}^{\mathrm{2}} \right)} \\ $$