TITLE: Sum of consecutive n integers
QUESTION [5 upvotes]: It is well-known that $1+2+3+4+...+n= \frac{n(n+1)}{2}$, this formula can be found using simple arithmetic progression. But the sum also can be found by using ${n+1 \choose 2}$, which is n+1 choose 2. Can anyone explain the relationship between the binomial coefficient and the sum of consecutive integers? I found the binomial coefficient formula at wikipedia: http://en.wikipedia.org/wiki/Triangular_number

REPLY [6 votes]: I think this is a better way that justifies the name Triangular Number. Consider the (right) triangular arrangement with $n$ rows and columns, the rows indexed by $i = 1, 2, \ldots n$, and the columns indexed by $0, 1, \ldots n - 1$:
$$\begin{matrix}1 \\ 2 \\ 3 \\ \vdots \\ n\\ \ \end{matrix}\left|
\begin{matrix}
*&\\
*& *\\
*& *& *&\\
\vdots& \vdots& \vdots& \ddots\\
*& *& *& \ldots & *\\
\hline
0 & 1 & 2 & \ldots & n-1
\end{matrix}\right.$$
The total number of objects is obviously $1 + 2 + \ldots + n$ (the sum of the number of objects in each row). However, observe that each object corresponds to a unique pair of the form $(i, j)$ with $j < i$, where $i$ is the row and $j$ is the column occupied by the object. Thus the number of objects is the number of such pairs, which is the number of distinct pairs of the $n + 1$ numbers $0, 1, \ldots, n$, and that is $\binom {n + 1} 2$.