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$S=\frac{1}{4(1)^{4}+1}+\frac{2}{4(2)^{4}+1}+\frac{3}{4(3)^{4}+1}+\dots+\frac{2016}{4(2016)^{4}+1}$

If the sum above $S = \dfrac{2017p}{2017q+1}$, where $p$ and $q$ are positive integers, find $p+q$.

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