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Question Number 154255 by EDWIN88 last updated on 16/Sep/21

(5/(3^2 .7^2 ))+(9/(7^2 .11^2 ))+((13)/(11^2 .15^2 ))+…=?

$$\:\frac{\mathrm{5}}{\mathrm{3}^{\mathrm{2}} .\mathrm{7}^{\mathrm{2}} }+\frac{\mathrm{9}}{\mathrm{7}^{\mathrm{2}} .\mathrm{11}^{\mathrm{2}} }+\frac{\mathrm{13}}{\mathrm{11}^{\mathrm{2}} .\mathrm{15}^{\mathrm{2}} }+\ldots=? \\ $$

Answered by ARUNG_Brandon_MBU last updated on 16/Sep/21

S=(5/(3^2 .7^2 ))+(9/(7^2 .11^2 ))+((13)/(11^2 .15^2 ))+… =Σ_(n≥1) ((4n+1)/((4n−1)^2 (4n+3)^2 ))=(1/8)Σ_(n≥1) ((1/((4n−1)^2 ))−(1/((4n+3)^2 ))) =(1/8)((1/3^2 )−(1/7^2 )+(1/7^2 )−(1/(11^2 ))+(1/(11^2 ))−(1/(15^2 ))+∙∙∙)=(1/8)×(1/3^2 )=(1/(72))

$${S}=\frac{\mathrm{5}}{\mathrm{3}^{\mathrm{2}} .\mathrm{7}^{\mathrm{2}} }+\frac{\mathrm{9}}{\mathrm{7}^{\mathrm{2}} .\mathrm{11}^{\mathrm{2}} }+\frac{\mathrm{13}}{\mathrm{11}^{\mathrm{2}} .\mathrm{15}^{\mathrm{2}} }+\ldots \\ $$$$\:\:\:=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{4}{n}+\mathrm{1}}{\left(\mathrm{4}{n}−\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{4}{n}+\mathrm{3}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{8}}\underset{{n}\geqslant\mathrm{1}} {\sum}\left(\frac{\mathrm{1}}{\left(\mathrm{4}{n}−\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\left(\mathrm{4}{n}+\mathrm{3}\right)^{\mathrm{2}} }\right) \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{8}}\left(\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{7}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{7}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{11}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{11}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{15}^{\mathrm{2}} }+\centerdot\centerdot\centerdot\right)=\frac{\mathrm{1}}{\mathrm{8}}×\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{72}} \\ $$

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