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Say I have n=5 I would like to perform the following equation:

5^2 + 4^2 + 3^2 + 2^2 + 1^2 = 55

I know that a triangular number can be worked out with the following equation:

(n(n+1)/2)

However it is at this point where I've not been able to work out the required adjustments to suite this use case.

What would be the correct equation?

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  • $\begingroup$ $$\sum_{i=1}^n i^{2}$$ $\endgroup$ – nissim abehcera Aug 24 '19 at 22:08
  • $\begingroup$ Hint: it's a cubic. $\endgroup$ – Peter Taylor Aug 24 '19 at 22:21
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    $\begingroup$ oeis.org/A000330 $\endgroup$ – rogerl Aug 24 '19 at 22:33
  • $\begingroup$ The general case is known as Faulhaber's formula $\endgroup$ – Ross Millikan Aug 24 '19 at 22:34
  • $\begingroup$ @RossMillikan thanks for the info! $\endgroup$ – jackdh Aug 24 '19 at 22:39
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$$\sum_{k=1}^n k^2 = \frac{1}{6}n(n+1)(2n+1)$$ with n=5 :$$\frac{1}{6}*5*6*11=55$$ it is sometimes called number for quadratic pyramid. trula

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