Finding $\sum\limits_{k=0}^n k^2$ using summation by parts Sorry to bother you guys again, but I still have some doubts. I do think I'm making some progress, though.
So, again, the formula that I'm using for summation by parts is

$\sum\limits_{k=o}^n f(k)g(k) = g(n)(\sum\limits_{k=0}^n f(k)) - \sum\limits_{k=0}^{n-1} [\Delta g(k) \sum\limits_{i=0}^k f(i)]$

Using this, I'm trying to find $\sum\limits_{k=0}^n k^2$. I think I almost got it, but there are some problems that I still can't get around.
Plugging in $g(k) = k$ and $f(k)=k$, we end up with the following equations:
\begin{align}
\sum\limits_{k=0}^n k^2 &= \sum\limits_{k=0}^n kk\\
\sum\limits_{k=0}^n k^2 &= n (\sum\limits_{k=0}^n k) - \sum\limits_{k=0}^{n-1}[\sum\limits_{i=0}^k i]\\
\sum\limits_{k=0}^n k^2 &= n(\frac{n(n+1)}{2}) - \sum\limits_{k=0}^{n-1}[\frac{k(k+1)}{2}]\\
\sum\limits_{k=0}^n k^2 &= \frac{n^3 + n^2}{2} - \frac{\sum\limits_{k=0}^{n-1} k^2+k}{2}\\
2\sum\limits_{k=0}^n k^2 &= n^3 + n^2 - \sum\limits_{k=0}^{n-1} k^2+k\\
2\sum\limits_{k=0}^n k^2 &= n^3 + n^2 - \sum\limits_{k=0}^{n-1} k^2+ \sum\limits_{k=0}^{n-1}k
\end{align}
Anyway, my problem is, I need the upper indexes of the sums in the right hand side to be $n$, not $n-1$ (I've already checked that, if the indexes are $n$, then I can find the desired result). Is there any trick I'm missing to make those indexes go up by $1$? Or am I missing an obvious step?

I dont want to be polemic but the formula you are using is a bit ugly... I like more this version
$$\sum f\Delta g=fg-\sum \Delta f\ \mathrm E g$$
Where delta is the difference operator, i.e. $\Delta f(n)=f(n+1)-f(n)$, and E is the shift operator, i.e. $\mathrm E f(n)=f(n+1)$.
I like to use too it recursive version when is possible, i.e., when you can represent an "analytic" difference or sum of a function
$$\sum f \Delta g=\sum_{k\ge 0}(-1)^k\ \Delta^k f\ \frac{\mathrm E^k}{\Delta^k}g$$
In your case using last formula I can write
$$f(n)=n \to \Delta f(n)=1 \to \Delta^k f(n)=0,\ k> 1\\
\Delta g(n)=n \to g(n)=\frac{n^\underline 2}{2} \to \frac{\mathrm E^k}{\Delta^k}g(n)=\frac{(n+k)^\underline {k+2}}{(k+2)!}$$
Where $n^\underline k$ is the falling factorial. You have 2 iteration because $\Delta^k f(n)=0,\ k> 1$. Then
$$\sum n^2 \delta n=n\frac{n^\underline 2}{2}-\frac{(n+1)^\underline 3}{6}$$
Taking limits you have
$$\sum_{n=0}^{N}n^2=\sum\nolimits_{0}^{N+1} n^2 \delta n=(N+1)\frac{(N+1)^\underline 2}{2}-\frac{(N+2)^\underline 3}{6}=\frac{1}{3}N^3+\frac{1}{2}N^2+\frac{1}{6}N$$
