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∑r=12013(1+r−2+(r+1)−2)12= ? \displaystyle \sum_{r=1}^{2013} ({1 + r^{-2} + (r+1)^{-2}})^{\frac{1}{2}} = \ ? r=1∑2013(1+r−2+(r+1)−2)21= ?
Give your answer to 3 decimal places.
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