Question
If $\lambda=\sum_{i=1}^{\infty} \frac{1}{i^{4}}$, then $\sum_{i=1}^{\infty} \frac{1}{(2 i-1)^{4}}$ is(A) $\frac{14}{15} \lambda$(B) $\frac{\lambda}{2}$(C) $\frac{16}{15} \lambda$(D) $\frac{15}{16} \lambda$
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