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If a=∑r=1∞1r2\displaystyle a=\sum_{r=1}^\infty \dfrac1{r^2}a=r=1∑∞r21 and b=∑r=1∞1(2r−1)2\displaystyle b=\sum_{r=1}^\infty \dfrac1{(2r-1)^2}b=r=1∑∞(2r−1)21, find 3ab\dfrac{3a}{b}b3a.
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