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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
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5 Answers

up vote 9 down vote accepted

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

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You just did their homework! –  Mitch Apr 1 '13 at 2:44
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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
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$k(k+1) = \frac{1}{3}((k+1)^3-k^3-1)$. 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
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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 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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