How to calculate the sum of the following series? $$(n-1)^2+(n-2)^2+...+1$$Thank you in advance
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1I'll google it for you: https://en.wikipedia.org/wiki/Square_pyramidal_number – gukoff Apr 28 '13 at 04:16
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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. – Antonio Vargas Apr 28 '13 at 04:17
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Here's a video explaining the solution. – vadim123 Apr 28 '13 at 04:17
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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. – Julien Apr 28 '13 at 04:19
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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 $. – Mhenni Benghorbal Apr 28 '13 at 04:29
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possible duplicate of How do I come up with a function to count a pyramid of apples?, also (2), (3), (4). – MJD Apr 28 '13 at 05:34
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Also (5), (6). – MJD Apr 28 '13 at 05:46
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@vadim123 Thank you very much – proofy Apr 28 '13 at 07:55
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@Harold Thank you very much – proofy Apr 28 '13 at 07:56
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@MJD Thank you very much – proofy Apr 28 '13 at 07:57
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@julien: Thanks for the comment. $\sum_{k=1}^{n-1}(n-k)^2 \implies n^2\sum_{k=1}^{n-1}1-2n \sum_{k=1}^{n-1}k + \sum_{k=1}^{n-1} k^2$. – Mhenni Benghorbal Apr 28 '13 at 10:08
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@Mhenni Benghorbal Thank you very much – proofy Apr 28 '13 at 10:57
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@proofy: You are welcome. – Mhenni Benghorbal Apr 28 '13 at 19:27
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I do not really understand why this was closed as too localized. This particular summation is certainly of interesting (although well known). That said I can see that this is an exact duplicate. – Baby Dragon Apr 28 '13 at 22:37
2 Answers
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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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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}