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Question Number 204642 by BaliramKumar last updated on 24/Feb/24
If (1/1^2 )+(1/2^2 )+(1/3^2 )+ (1/4^2 )+(1/5^2 ) + ............. = (π^2 /6) then (1/1^2 )+(1/3^2 )+(1/5^2 ) + ............. = ?
$$\mathrm{If}\:\:\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\:\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} }\:+\:………….\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\mathrm{then}\:\:\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} }\:+\:………….\:=\:? \\ $$$$ \\ $$
Answered by Rasheed.Sindhi last updated on 24/Feb/24
(1/1^2 )+(1/2^2 )+(1/3^2 )+ (1/4^2 )+(1/5^2 ) + ............. = (π^2 /6) ((1/1^2 )+(1/3^2 )+(1/5^2 ) +∙∙∙)+( (1/2^2 )+ (1/4^2 )+(1/6^2 ) +∙∙∙)=(π^2 /6) ((1/1^2 )+(1/3^2 )+(1/5^2 ) +∙∙∙)+(1/2^2 )( (1/1^2 )+ (1/2^2 )+(1/3^2 ) +∙∙∙)=(π^2 /6) ((1/1^2 )+(1/3^2 )+(1/5^2 ) +∙∙∙)+(1/4)( (π^2 /6))=(π^2 /6) ((1/1^2 )+(1/3^2 )+(1/5^2 ) +∙∙∙)=(π^2 /6)−(π^2 /(24))=((4π^2 −π^2 )/(24))=(π^2 /8)
$$\:\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\:\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} }\:+\:………….\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} }\:+\centerdot\centerdot\centerdot\right)+\left(\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\:\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{2}} }\:+\centerdot\centerdot\centerdot\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} }\:+\centerdot\centerdot\centerdot\right)+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\left(\:\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\:+\centerdot\centerdot\centerdot\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} }\:+\centerdot\centerdot\centerdot\right)+\frac{\mathrm{1}}{\mathrm{4}}\left(\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} }\:+\centerdot\centerdot\centerdot\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\frac{\pi^{\mathrm{2}} }{\mathrm{24}}=\frac{\mathrm{4}\pi^{\mathrm{2}} −\pi^{\mathrm{2}} }{\mathrm{24}}=\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$
Commented by BaliramKumar last updated on 24/Feb/24
Thanks
$$\mathrm{Thanks} \\ $$

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