Sum of the first $n$ quadratic integers

Question

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.

Answer 1

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, $$\sum_{k=1}^{n} k^{\ell_0} $$

Answer 2

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 $\sum_{k=1}^{n}k^3$.