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How to calculate the sum of the following series? $$(n-1)^2+(n-2)^2+...+1$$Thank you in advance

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closed as too localized by Antonio Vargas, 23rd, MJD, Andrés E. Caicedo, Learner Apr 28 '13 at 6:44

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    $\begingroup$ I'll google it for you: en.wikipedia.org/wiki/Square_pyramidal_number $\endgroup$ – gukoff Apr 28 '13 at 4:16
  • $\begingroup$ What have you tried? There are a few ways to find this sum, so if you describe your approach, where you are stuck, and any similar problems you have worked on then it will be more likely that you get an answer that is especially tailored to your needs. As it is, this question does not give any context and does not meet the quality standards here on Math.SE. $\endgroup$ – Antonio Vargas Apr 28 '13 at 4:17
  • $\begingroup$ Here's a video explaining the solution. $\endgroup$ – vadim123 Apr 28 '13 at 4:17
  • $\begingroup$ This is a finite sum, rather than a series (even if it every finite sum is a series...). And what you are looking for is a Faulhaber's formula. $\endgroup$ – Julien Apr 28 '13 at 4:19
  • $\begingroup$ A related problem. Note that, $\sum_{k=1}^{n-1}(n-k)^2 \implies n^2\sum_{k=1}^{n-1}1-2n \sum_{k=1}^{n-1}k + \sum_{k=1}^{n-1} 1 $. $\endgroup$ – Mhenni Benghorbal Apr 28 '13 at 4:29
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Hint:

\begin{align*} \sum\limits_{k=1}^n k^2 &= \frac{n(n+1)(2n+1)}{6}\\ \implies \sum_{k=1}^{n-1}k^2&= \hspace{1mm}? \end{align*}

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    $\begingroup$ This ISN'T an answer. $\endgroup$ – Squirtle Apr 28 '13 at 4:29
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    $\begingroup$ Looks like an answer to me. $\endgroup$ – MJD Apr 28 '13 at 5:51
  • $\begingroup$ @pedrosuavo Thank you very much $\endgroup$ – proofy Apr 28 '13 at 8:29
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First recall the following: $$\sum_{k=1}^{n-1} 1 = n-1$$ $$\sum_{k=1}^{n-1} k = \dfrac{n(n-1)}2$$ Now we have the identity $$(k+1)^3 - k^3 = 3k^2 + 3k + 1$$ Write this for $k=1,2,\ldots, n-2$ and we get \begin{align} 2^3-1^3 & = 3 \cdot 1^2 + 3 \cdot 1 + 1\\ 3^3-2^3 & = 3 \cdot 2^2 + 3 \cdot 2 + 1\\ 4^3-3^3 & = 3 \cdot 3^2 + 3 \cdot 3 + 1\\ \vdots & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \vdots\\ n^3-(n-1)^3 & = 3 \cdot (n-1)^2 + 3 \cdot (n-1) + 1\\ \end{align} Add these up and note that there is a telescopic cancellation on the left hand side to give $$n^3 - 1^3 = 3 \sum_{k=1}^{n-1} k^2 + 3 \sum_{k=1}^{n-1} k + \sum_{k=1}^{n-1} 1$$ Hence, we get that \begin{align} \sum_{k=1}^{n-1} k^2 & = \dfrac{n^3 - 1}3 -\sum_{k=1}^{n-1} k - \dfrac{\sum_{k=1}^{n-1} 1}3 = \dfrac{n^3 -1}3 - \dfrac{n(n-1)}2 - \dfrac{n-1}3\\ & = \dfrac{n^3-n}3 - \dfrac{n(n-1)}2 = \dfrac{n(n-1)(2n-1)}6 \end{align}

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