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If ∑n=1∞1n2=π26\displaystyle \sum_{n=1}^\infty \dfrac1{n^2} = \dfrac{\pi^2}6 n=1∑∞n21=6π2, then what is the value of ∑n=1∞1(2n−1)2 \displaystyle \sum_{n=1}^\infty \dfrac1{(2n-1)^2} n=1∑∞(2n−1)21?
Give your answer to 3 decimal places.
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