How to convert this sum into an equation?

$$ \sum_{i=1}^{C-1} i(R-i)(C-i) $$


$$ \sum_{i=1}^{C-1} (RCi - Ri^2 -Ci^2 + i^3) $$

Which can be represented as:

$$ RC\sum_{i=1}^{c-1}i - R\sum_{i=1}^{c-1}i^2-C\sum_{i=1}^{c-1}i^2 + \sum_{i=1}^{c-1}i^3 $$

Using the Faulhaber formula and having n = c-1, it can be represented as:

$$ {1\over 2}RC(n^2+n) -{1\over6}R(2n^3+3n^2+n) - {1\over 6}C(2n^3+3n^2+n)+{1\over4}(n^4+2n^3+n^2) $$

But the results from the sum and the formula from the Faulhaber formula are different. Where is my error?


closed as off-topic by Leucippus, JonMark Perry, José Carlos Santos, Brian Borchers, user223391 Mar 25 '18 at 0:12

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  • 2
    $\begingroup$ You mean, simplify this ? $\endgroup$ – Shailesh Mar 21 '18 at 2:59
  • $\begingroup$ I don't see any error in what you did. Maybe your sum is incorrect? $\endgroup$ – Ross Millikan Mar 21 '18 at 4:24
  • $\begingroup$ You are correct. There was an error in my sum. Marked your answer as correct since I wasn't sure it could be solved using Faulhaber's formula. $\endgroup$ – smash87 Mar 21 '18 at 4:57

Expand the product, which will give terms in $i, i^2,$ and $i^3$. Do you know how to sum each of those by Faulhaber's formula?

  • $\begingroup$ I tried using this but I don't get the same result. $\endgroup$ – smash87 Mar 21 '18 at 3:53
  • $\begingroup$ The same result as what? Yes, that is what you are looking for. If we don't see your work we can't find the problem. $\endgroup$ – Ross Millikan Mar 21 '18 at 3:55
  • $\begingroup$ I just edited the question to add the procedure $\endgroup$ – smash87 Mar 21 '18 at 4:11

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