<s>[INST] 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?
 [/INST] As you probably know:
$$\sum_{k=0}^{n-1} k = \frac{(n-1)^2+(n-1)}{2} = \frac{n^2-2n+1+n-1}{2} = \frac{n^2-n}{2}$$
You also know:
$$\sum_{k=0}^{n}k^2=n^2+\sum_{k=0}^{n-1}k^2 \rightarrow \sum_{k=0}^{n-1} k^2=\sum_{k=0}^{n} k^2 - n^2$$
So now you can substitute:
$$2\sum_{k=0}^n k^2=n^3+n^2-\left(\sum_{k=0}^n k^2-n^2\right)-\frac{n^2-n}{2}$$
Pass this over and you are good to go!
$$3\sum_{k=0}^n k^2=n^3+n^2+n^2-\frac{n^2-n}{2}$$
$$\sum_{k=0}^n k^2=\frac{n^3+\frac{3}{2}n^2-\frac{n}{2}}{3}=\frac{1}{3}n^3+\frac{1}{2}n^2+\frac{1}{6}n$$
</s>