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If ∑n=120161nn+1+(n+1)n=a−1a,\large \displaystyle \sum_{n=1}^{2016} \dfrac{1}{n\sqrt{n+1}+(n+1)\sqrt{n}} = \dfrac{\sqrt{{a}}-1}{\sqrt{{a}}},n=1∑2016nn+1+(n+1)n1=aa−1, find aaa.
Inspiration.
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