Using the method of differences, prove that $\displaystyle \sum_{r = 1}^{2n}r^2=\frac{1}{6}n(n+1)(2n+1)$

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- May 20th 2011, 08:34 PMPunchMethod of Differences
Using the method of differences, prove that $\displaystyle \sum_{r = 1}^{2n}r^2=\frac{1}{6}n(n+1)(2n+1)$

- May 20th 2011, 10:16 PMProve It
I don't know what the method of differences is, but

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This should also help you to get the required result...

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