<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] Divide each term of the series by $2$. The result is
$$\binom{2}{2}+\binom{3}{2}+\cdots+\binom{n+1}{2}.\tag{$1$}$$
We give a combinatorial argument that the sum $(1)$ is equal to $\binom{n+2}{3}$.
Now how many ways are there to choose three numbers  from the numbers $1$ to $n+2$? The smallest number chosen could be $n$. Then there are $\binom{2}{2}$ ways to choose the other two. Or the smallest chosen number could be $n-1$, in which case there are $\binom{3}{2}$ ways to choose the other two. Or the smallest chosen number could be $n-2$, in which case there are $\binom{4}{2}$ ways to choose the other two. And so on, up to the smallest chosen number being $1$, in which case there are $\binom{n+1}{2}$ ways to choose the other two. 
Thus half our sum is $\binom{n+2}{3}$, and we arrive at
$$1\cdot 2+2\cdot 3+\cdots+n\cdot(n+1)=2\binom{n+2}{3}.$$  
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