Formula for the $1\cdot 2 + 2\cdot 3 + 3\cdot 4+\ldots + n\cdot (n+1)$ sum

Solution 1:

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 $$1\cdot 2+2\cdot 3+\cdots+n\cdot(n+1)=2\binom{n+2}{3}.$$

Solution 2:

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

Solution 3:

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

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

Solution 4:

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.