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 Feb 24, 2019
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\(\sum \limits_{k=100}^1~(-1)^kk^2 = \\ (100^2-99^2) + (98^2-97^2) + \dots +(2k^2 - (2k-1)^2) + \dots + (2^2-1^2)=\\ \sum \limits_{k=50}^1~(2k)^2-(2k-1)^2 = \sum \limits_{k=50}^1~(4k-1) = \\ 4\cdot \dfrac{(50)(51)}{2}-50 = 5050\)

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 Feb 24, 2019

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