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?

$S_n = 1\cdot2 + 2\cdot 3 + 3\cdot 4 + 4\cdot 5 +\ldots + n\cdot (n+1)$

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Of course, a finite sum is a series. But it is first a finite sum. – 1015 Mar 31 '13 at 20:26
Here are techniques for summing the series. – Mhenni Benghorbal Mar 31 '13 at 20:30

$S_n = \sum_{k=1}^n k(k+1) = \sum_{k=1}^n k^2 + \sum_{k=1}^n k = \frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2} = \frac{n(n+1)(n+2)}{3}$

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You just did their homework! – Mitch Apr 1 '13 at 2:44

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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Nice alternative, +1. – 1015 Mar 31 '13 at 21:14

$k(k+1) = \frac{1}{3}((k+1)^3-k^3-1)$. So all the $k$'s cancel, except the first and last. We get:

$\sum_1^n k(k+1) = \frac{1}{3}\sum_1^n ((k+1)^3-k^3-1) = \frac{1}{3}((n+1)^3-n-1) = \frac{1}{3}n(n+1)(n+2)$

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Clever...but how did you do the magic to think of converting $k(k+1)$ to the inner two terms of a cubed binomial? – Mitch Apr 1 '13 at 2:47

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.

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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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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 $S_n$.

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