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sum(i, 1, n, i^2) = n(n+1)(2n+1)/6

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Calculus & discrete math

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\sum_{i=1}^{n} i^{2} = \frac{\operatorname{n}\left(n + 1\right) \cdot \left(2 \cdot n + 1\right)}{6}

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Unknown function "n"