Sum of the first $n$ quadratic integers 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.
 A: 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 $\sum_{k=1}^{n}k^3$.
A: 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,
$$\sum_{k=1}^{n} k^{\ell_0}  $$
