Averages/Root mean square: Difference between revisions
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→Not using a loop
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Using these equations:<br>
<math>\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}</math> See [[wp:List of mathematical series]]<br><br>
for <math>
We can show that:<br>
<math>\sum_{i=a}^b i^2 = \frac{b(b+1)(2b+1)-a(a-1)(2a-1)}{6}</math>
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