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Given that: $$\sum_{n=1}^\infty \frac{1}{n^4}= \frac{\pi^4}{90}$$ Find the sum of: $$\sum_{n=0}^\infty \frac{1}{(2n+1)^4}$$

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We have that

$$\sum_{n=0}^\infty \frac{1}{(2n+1)^4}=\sum_{n=1}^\infty \frac{1}{n^4}-\sum_{n=1}^\infty \frac{1}{(2n)^4}$$

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It is given that ${\sum_{n=1}^{\infty}}{1\over {n^4}}={{\pi}^2\over 90}$

We have to find $${\sum_{n=0}^{\infty}}{1\over {(2n+1)^4}}$$ $$={\sum_{n=0}^{\infty}}{1\over {n^4}}-{1\over {(2n)^4}}$$ $$={\sum_{n=0}^{\infty}}{1\over {n^4}}-{1\over 16}{\sum_{n=0}^{\infty}}{1\over {n^4}}$$ $$={{\pi}^2\over 90}\times {15\over 16}$$ $$={{\pi}^2\over 96}$$

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