If sumlimits_{r=1}^infty frac{1}{(2r-1)^2} = | vedclass

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(C) Let $S = \sum\limits_{r=1}^\infty \frac{1}{r^2} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \dots$We can split this into odd

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$\frac{\pi^2}{6}$

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(C) Let $S = \sum\limits_{r=1}^\infty \frac{1}{r^2} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \dots$We can split this into odd and even terms:$S = \left( \frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \dots ight) + \left( \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \dots ight)$Given $\sum\limits_{r=1}^\infty \frac{1}{(2r-1)^2} = \frac{\pi^2}{8}$,the sum of odd terms is $\frac{\pi^2}{8}$.The sum of even terms is $\sum\limits_{r=1}^\infty \frac{1}{(2r)^2} = \frac{1}{4} \sum\limits_{r=1}^\infty \frac{1}{r^2} = \frac{1}{4} S$.Therefore,$S = \frac{\pi^2}{8} + \frac{1}{4} S$.$S - \frac{1}{4} S = \frac{\pi^2}{8} \implies \frac{3}{4} S = \frac{\pi^2}{8}$.$S = \frac{\pi^2}{8} 	imes \frac{4}{3} = \frac{\pi^2}{6}$.

If $\sum\limits_{r=1}^\infty \frac{1}{(2r-1)^2} = \frac{\pi^2}{8}$,then $\sum\limits_{r=1}^\infty \frac{1}{r^2} = \dots$

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