Find a formula for $0 · 1 · 2 + 1 · 2 · 3 + 2 · 3 · 4 + \dots +n(n + 1)(n + 2)$, for $n \in \mathbb N$ $$\sum\limits_{i=1}^n i(i + 1)(i + 2)$$
$$\sum\limits_{i=1}^n i^3 + 3i^2 + 2i$$
$$\sum\limits_{i=1}^n i^3 + 3\sum\limits_{i=1}^ni^2 + 2\sum\limits_{i=1}^ni$$
$$= (\frac14)n^4 + (\frac12)n^3 + (\frac14)n^2 + n^3 + 3(\frac{n^2}2) + (\frac{n}2) + n^2 + n$$
$$ =n^4 + 3n^3 + 8n^2 + 3n$$
Why does what I did above not work?
 A: It is very much easier to note that $$n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)=4n(n+1)(n+2)$$ from which you get a telescoping series.
This is related to the binomial identity $\binom nr+\binom n{r+1}=\binom {n+1}{r+1}$ with $r=3$, and the pattern clearly generalises.
A: Required formulas:
$$\sum_{i=1}^n i^3=\left(\frac{n(n+1)}2\right)^2$$
$$\sum_{i=1}^n i^2=\frac{n(n+1)(2n+1)}{6}$$
$$\sum_{i=1}^n i=\frac{n(n+1)}2$$

I will start off at the third step.
$$\sum_{i=1}^n i^3+3\sum_{i=1}^n i^2+2\sum_{i=1}^n i$$
$$=\left(\frac{n(n+1)}{2}\right)^2+3\cdot \frac{n(n+1)(2n+1)}{6}+2\cdot \frac{n(n+1)}{2}$$
$$=\frac{n^2(n+1)^2}{4}+\frac{n(n+1)(2n+1)}{2}+n(n+1)$$
$$=\frac{n^2(n^2+2n+1)}{4}+\frac{2n(2n^2+3n+1)}{4}+\frac{4n^2+4n}{4}$$
$$=\frac{n^4+2n^3+n^2+4n^3+6n^2+2n+4n^2+4n}{4}$$
$$=\frac{n^4+6n^3+11n^2+6n}{4}$$
$$\color{green}{\sum_{i=1}^n i(i+1)(i+2)=\frac{n^4+6n^3+11n^2+6n}{4}}$$
A: In addition to these excellent answers, there is a general approach that words for all partial sums of the form:
$$S(n) = \sum_{k=1}^n f(k),$$
where $f$ is a degree $d$ polynomial.
We must have that $S(n) = g(n)$, where $g$ is a degree $d+1$ polynomial, with $d+2$ coefficients to determine.  Then we evaluate $S(n)$, for $n=1,\dots,d+2$, to get a linear system with $d+2$ equations and unknowns that can be solved for.
More strikingly, if you have a proposed solution to such a problem, a proof of its correctness requires merely that you check its correctness for $d+2$ different values of $n$!
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\sum_{k = 1}^{n}k\pars{k + 1}\pars{k + 2}:\ {\large ?}}$

\begin{align}&\color{#c00000}{%
\sum_{k = 1}^{n}k\pars{k + 1}\pars{k + 2}}
=\sum_{k = 1}^{n}{\pars{k + 2}! \over \pars{k - 1}!}
=3!\sum_{k = 1}^{n}{k + 2 \choose 3}
=6\sum_{k = 1}^{n}\oint_{\verts{z}=1}{\pars{1 + z}^{k + 2} \over z^{4}}
\,{\dd z \over 2\pi\ic}
\\[3mm]&=6\oint_{\verts{z}=1}{1 \over z^{4}}\sum_{k = 1}^{n}\pars{1 + z}^{k + 2} 
\,{\dd z \over 2\pi\ic}
=6\oint_{\verts{z}=1}{1 \over z^{4}}\,
{\pars{1 + z}^{3}\bracks{\pars{1 + z}^{n} - 1} \over \pars{1 + z} - 1} 
\,{\dd z \over 2\pi\ic}
\\[3mm]&=6\ \underbrace{\oint_{\verts{z}=1}{\pars{1 + z}^{n + 3} \over z^{5}}
\,{\dd z \over 2\pi\ic}}_{\ds{=\ {n + 3 \choose 4}}}\ -6\
\underbrace{\oint_{\verts{z}=1}{\pars{1 + z}^{3} \over z^{5}}
\,{\dd z \over 2\pi\ic}}_{\ds{=\ 0}}\ =\ 6{n + 3 \choose 4}
\\[3mm]&=6\,{\pars{n + 3}\pars{n + 2}\pars{n + 1}n \over 4\cdot 3\cdot 2\cdot 1}
\end{align}

$$\color{#00f}{\large%
\sum_{k = 1}^{n}k\pars{k + 1}\pars{k + 2}
={1 \over 4}\,\pars{n + 3}\pars{n + 2}\pars{n + 1}n}
$$
A: $$
0 \cdot 1 \cdot 2 + 1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + \cdots +n(n + 1)(n + 2)
=
3!\ \sum_{k=1}^{n+2} \binom{k}{3}
=
6 \binom{n+3}{4} 
$$
The key sum is the sum of a column of Pascal's triangle, which is found by induction using Pascal's rule.
