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If ∑r=1∞rr4+r2+1=1−1a \displaystyle \sum_{r=1}^\infty \dfrac r{r^4+r^2+1} = 1 - \dfrac1a r=1∑∞r4+r2+1r=1−a1, find the number of digits in a2016a^{2016} a2016.
You are given the following approximations: log10=1,log5=0.6990,log3=0.4771\log 10 = 1, \log 5 = 0.6990, \log 3 = 0.4771 log10=1,log5=0.6990,log3=0.4771.
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