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

Let $S$ be our sum.  Then
$$S=\binom{2}{2}+\binom{3}{2}+\binom{4}{2} + \cdots + \binom{2012}{2}=\binom{2013}{3}=\frac{2013\cdot 2012\cdot 2011}{3 \cdot 2 \cdot 1}.$$
Justification:  We count, in two different ways, the number of ways to choose $3$ numbers from the set 
$$\{1,2,3,4,\dots, n,n+1\}.$$
(For our particular problem we use $n=2012$.)
First Count: It is clear that there are $\binom{n+1}{3}$ ways to choose $3$ numbers from $n+1$ numbers.  
Second Count: The smallest chosen number could be $1$.  Then there are $\binom{n}{2}$ ways to choose the remaining $2$ numbers. 
Or the smallest chosen number could be $2$, leaving $\binom{n-1}{2}$ choices for the remaining  $2$ numbers.  Or the smallest chosen number could be $3$, leaving $\binom{n-2}{2}$ choices for the remaining $2$ numbers.  And so on, up to smallest chosen number being $n-1$, in which case there are $\binom{2}{2}$ ways to choose the remaining $2$ numbers. Thus the total count is
$$\binom{n}{2}+\binom{n-1}{2}+\binom{n-2}{2}+\cdots +\binom{3}{2}+\binom{2}{2}.$$
Comparing the two counts, we find that 
$$\binom{2}{2}+\binom{3}{2}+\binom{4}{2}+\cdots +\binom{n-1}{2}+\binom{n}{2}=\binom{n+1}{3}.$$
Comment: Similarly, it is easy to see that in general $\sum_{k=r}^n \binom{k}{r}=\binom{n+1}{r+1}.$  These natural binomial coefficient identities give a combinatorial approach to finding general formulas for the sums of consecutive squares, consecutive cubes, and so on.
