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For all positive integers n\large nn,
12+22+32+⋯+n2=n(n+1)(2n+1)6\large 1^2 + 2^2 + 3^2 + \dots + n^2 = \frac {n(n + 1)(2n + 1)}{6}12+22+32+⋯+n2=6n(n+1)(2n+1)
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