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__KevinMZ__
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发布时间
2024/8/14 12:43
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2024/8/14 14:46:31
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1~4幂次方和
__KevinMZ__
楼主
2024/8/14 12:43
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\sum_{i = 1}^n i^0 = n \\ \sum_{i = 1}^n i^1 = \frac{n(n + 1)}{2} \\ \sum_{i = 1}^n i^2 = \frac{n(n + 1)(2n + 1)}{6} \\ \sum_{i = 1}^n i^3 = (\frac{n(n + 1)}{2})^2 \\ \sum_{i = 1}^n i^4 = \frac{n(n + 1)(2n + 1)(3n^2 + 3n - 1)}{30}
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2024/8/14 12:43
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