$\displaystyle \sum_{i=1}^n i^2 = \frac{1}{6}n(n+1)(2n+1) $

edit , it works now,

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- May 23rd 2012, 07:45 AMTweetylatex help, why does this formula not work?
$\displaystyle \sum_{i=1}^n i^2 = \frac{1}{6}n(n+1)(2n+1) $

edit , it works now,