How can the following be calculated? How can the following series be calculated?
$$S=1+(1+2)+(1+2+3)+(1+2+3+4)+\cdots+(1+2+3+4+\cdots+2011)$$

If you look at the third differences then you will get a constant, so you know there is a third degree polynomial formula.  In fact, since the constant is 1, you also know the coefficient of $n^3$ is $\frac{1}{3!}$.
So just fit the early terms $S_0=0, S_1=1, S_2=4, S_4=10$ to a third degree polynomial and find the result $$S_n = \frac{n^3}{6} +\frac{n^2}{2} +\frac{n}{3}.$$
Then let $n=2011$.
