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?
$S_n = 1\cdot2 + 2\cdot 3 + 3\cdot 4 + 4\cdot 5 +\ldots  + n\cdot (n+1)$
 A: $k(k+1) = \frac{1}{3}((k+1)^3-k^3-1)$. So all the $k$'s cancel, except the first and last. We get:
$\sum_1^n k(k+1) = \frac{1}{3}\sum_1^n ((k+1)^3-k^3-1) = \frac{1}{3}((n+1)^3-n-1) = \frac{1}{3}n(n+1)(n+2)$
A: Not a smart way, but it is well-known that we have
$$
\sum_{k=1}^nk=\frac{n(n+1)}{2}\qquad\mbox{and}\qquad \sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6}.
$$
So
$$
\sum_{k=1}^nk(k+1)=\sum_{k=1}^nk^2+k=\sum_{k=1}^nk^2+\sum_{k=1}^nk=\ldots
$$
TonyK's answer is highly recommended: that's the smart way.
A: 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}
A: 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
$$1\cdot 2+2\cdot 3+\cdots+n\cdot(n+1)=2\binom{n+2}{3}.$$  
A: $S_n = \sum_{k=1}^n k(k+1) =  \sum_{k=1}^n k^2 +  \sum_{k=1}^n k = \frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2} = \frac{n(n+1)(n+2)}{3}$
A: We have
$$ S_n = \sum_{k=1}^n k(k+1) = \sum_{k=1}^n k^2 + \sum_{k=1}^n k.$$
There are formulas for both the sum of the first $n$ natural numbers and the sum of their squares, which leads to a simple formula for $S_n$.
