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
$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{9}\left({n}+\mathrm{1}\right)×\mathrm{3}^{{n}} ×\left(\sqrt{{u}}\right)^{{n}} }{\left({x}^{{n}} +{y}^{{n}} \right)^{\mathrm{2}} \left(\sqrt[{\mathrm{3}}]{{z}}\right)^{{n}} } \\ $$$$=\mathrm{9}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left({n}+\mathrm{1}\right)}{\left({x}^{{n}} +{y}^{{n}} \right)^{\mathrm{2}} }\left(\frac{\mathrm{3}\sqrt{{u}}}{\:\sqrt[{\mathrm{3}}]{{z}}}\right)^{{n}} \\ $$$$=\mathrm{9}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left({n}+\mathrm{1}\right)}{\left({a}^{{n}} +{b}^{{n}} \right)^{\mathrm{2}} }\:{with}\:{a}={x}\sqrt{\frac{\sqrt[{\mathrm{3}}]{{z}}}{\mathrm{3}\sqrt{{u}}}},\:{b}={y}\sqrt{\frac{\sqrt[{\mathrm{3}}]{{z}}}{\mathrm{3}\sqrt{{u}}}} \\ $$$$……. \\ $$
Commented by mr W last updated on 29/Aug/24
$${unsolvable}\:{or}\:{solvable}? \\ $$