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I'm from Italy so maybe my English could be bad. To prove that $$ S=\sum_{k=1}^{n} k^2 = \frac{n(2n+1)(n+1)}{6} $$
we can consider the series: $$ \sum_{k=1}^{n}\left((k-1)^3-k^3\right) \, . $$ Note that \begin{align} (k-1)^3 &= k^3 -1 -3k^2 +3k \\[5pt] (k-1)^3 - k^3 &= -1 -3k^2 +3k \end{align} and so \begin{align} \sum_{k=1}^{n}((k-1)^3 - k^3) &= \sum_{k=1}^{n}(-1 -3k^2 +3k) \\[5pt] -n^3 &= -n -3S + \frac{3n(n+1)}{2} \end{align} etc etc

I understand why this proof works, but I dont understand the initial thought to use $(k-1)^3$ and same for $\sum_{k=1}^{n} k^3$ we can start from $(k-1)^4$ and so on. Which reason there is behind that?

I hope to have been clear. Thanks for the attention.

amWhy
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Edward
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  • You want to know why the sum of $k^2$, but you use expressions cubed???? – amWhy Jun 08 '21 at 19:30
  • $\sum_{k=1}^n k^2= 1^2 + 2^2 + 3^2 ....$ – amWhy Jun 08 '21 at 19:33
  • @amWhy: In this thread, it is explained how you can use $\sum_{k=1}^{n}\left((k+1)^3-k^3\right)$ to find the sum of $n$ square numbers. – Joe Jun 08 '21 at 19:33
  • Then please include that link in the question post on behalf of Edward, @Joe. – amWhy Jun 08 '21 at 19:35
  • @amWhy: It's done. Now I look at things more closely, Edward considers the series $\sum_{k=1}^{n}\left((k-1)^3-k^3\right)$ but the idea is obviously the same. – Joe Jun 08 '21 at 19:44
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    Incidentally, the usual term for the values $k^2$ for natural numbers $k$ is "squares" (or "perfect squares"). So the title of your post would more typically be written as "Sum of the first $n$ (positive) squares." – Brian Tung Jun 08 '21 at 19:53

2 Answers2

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In mathematics, good things happen when you can evaluate something in two different ways. The telescoping series $$ S_1=\sum_{k=0}^{n}(k-1)^3-k^3 \, , $$ can be evaluated using the 'method of differences', where most of the terms are cancelled out. This tells us that $$ S_1=-n^3 \, . $$ However, we also know that $$ S_1=\sum_{k=0}^{n}-3k^2+3k-1 \tag{*}\label{*} \, . $$ which expresses $S_1$ in terms of the unknown $S=\sum_{k=1}^{n}k^2$. Because we already know that $S_1=-n^3$ using the method of differences, we can rearrange $\eqref{*}$ to make $S$ the subject of the equation and solve.

The same principle applies when computing series such as $\sum_{k=1}^{n}k^3$: because we can evaluate $\sum_{k=1}^{n}(k-1)^4-k^4$ in two different ways, this gives us enough information to find $\sum_{k=1}^{n}k^3$.

Joe
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  • Look, u just repeat what I know yet. Imagine a world where the telescoping series formula wasnt discovered yet; imagine you are the one that should proof the formula I asked; how can u think to adjust your $k^a$ equation by start from $(k-1)^{a+1}$? – Edward Jun 09 '21 at 01:45
  • @Edward: It makes a lot of sense to find a sum that you already know the value of which can be expressed in terms of the unknown $\sum k^a$. One way to find a sum that you already of is to consider other summation methods, such as the 'method of differences'. I'm sorry if this isn't the kind of answer that you're looking for. – Joe Jun 09 '21 at 08:17
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This is a more general answer than what you are actually asking.

The reason is for me as follows: you only know the sums for $\ell< \ell_0$, and you want to calculate $$\sum_{k=1}^{n} k^{\ell_0} $$

The terms in $(k+1)^{\ell_0+1}-k^{\ell_0+1}$ are all of the form $\alpha_{\ell} k^{\ell}$ with $\ell \leq \ell_0$ so you know how to calculate each $$\alpha_{\ell}\displaystyle \sum_{k=1}^n k^{\ell} \quad \forall \ell<\ell_0$$

Except $$\alpha_{\ell_0}\sum_{k=1}^{n} k^{\ell_0} $$

Moreover, $\sum_{k=1}^n \left((k+1)^{\ell_0+1}-k^{\ell_0+1}\right) $ is a telescoping sum so it is really easy to calculate.

You can see that $$\sum_{k=1}^n \left((k+1)^{\ell_0+1}-k^{\ell_0+1}\right) = (n+1)^{\ell_0+1}-1 $$

Therefore it gives you enough information to determine, $$\sum_{k=1}^{n} k^{\ell_0} $$

Axel
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