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Question Number 211129 by depressiveshrek last updated on 29/Aug/24

Commented by depressiveshrek last updated on 29/Aug/24

$$\mathrm{Find}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{this}\mathrm{series}$$

Answered by mr W last updated on 29/Aug/24

$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{9(n+1)\times 3^n\times {\left(\sqrt{u}\right)}^n}{{(x^n+y^n)}^2{\left(\sqrt[3]{z}\right)}^n}$$$$=9\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{(n+1)}{{(x^n+y^n)}^2}{\left(\frac{3\sqrt{u}}{\sqrt[3]{z}}\right)}^n$$$$=9\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{(n+1)}{{(a^n+b^n)}^2}witha=x\sqrt{\frac{\sqrt[3]{z}}{3\sqrt{u}}},b=y\sqrt{\frac{\sqrt[3]{z}}{3\sqrt{u}}}$$$$.......$$

Commented by mr W last updated on 29/Aug/24

$$unsolvableorsolvable?$$

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