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∑n=1101(2n+112+22+32+⋯+n2)\large \displaystyle \sum_{n=1}^{101} \left( \dfrac{2n+1}{1^2 + 2^2 + 3^2 + \cdots + n^2} \right )n=1∑101(12+22+32+⋯+n22n+1)
If the value of above expression can be written in the form AB\dfrac{A}{B}BA, where AAA and BBB are positive coprime integers, find A+BA+BA+B.
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