<s>[INST] Formula for the $1\cdot 2 + 2\cdot 3 + 3\cdot 4+\ldots + n\cdot (n+1)$ sum Is there a formula for the following sum?
$S_n = 1\cdot2 + 2\cdot 3 + 3\cdot 4 + 4\cdot 5 +\ldots  + n\cdot (n+1)$
 [/INST] We can use discrete calculus!  Let $x^{\overline{k}}$ denote the $k$th rising factorial power of $x$.  That is: $$x^{\overline k} = \underbrace{x(x+1)(x+2)\cdots(x+k-1)}_{k\text{ factors}}$$  
Then, $S_n = \sum_{k=0}^{n} k(k+1) = \sum_0^{n+1}x^{\overline 2}\,\delta x$.  (Notice that I started the summation at $k=0$; this makes it easier to plug in the lower limit, but does not affect the value of the summation since the first term is $0$.)
Using the power rule for summation, we have:
\begin{align}
S_n &= \sum_0^{n+1}x^{\overline 2}\,\delta x\\
&= \frac{(x-1)^{\overline 3}}{3}\Bigg|_0^{n+1}\\
&= \frac{n^{\overline 3}}{3}\\
&= \frac{n(n+1)(n+2)}{3}
\end{align}
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