Is there a combinatorial interpretation of the triangular numbers? The triangular numbers count the number of items in a triangle with $n$ items on a side, like this:

This can be calculated exactly by the formula $T_n = \sum_{k=1}^n k = \frac{n(n+1)}{2} = {n+1 \choose 2} = {n+1 \choose n-1}$.
Is there any combinatorial interpretation to that formula, as in some way to interpret arranging objects in a triangle with $n$ on a side as the number of ways to choose 2 or $n-1$ objects out of a collection of $n+1$ objects?

Here is a combinatorial proof of the identity
$$
1+2+\dotsb+n=\binom{n+1}{2}.
$$
The RHS counts the number of two-element subsets of $\{0,1,\dotsc,n\}$. Let $S _k$ be those two-element subsets of the preceding set with larger element $k$ for $k=1,\dotsb, n$. Then the $S _k$ partition the set of two element subsets of $\{0,1,\dotsc,n\}$. Further, $|S _k|=k$. Counting in this way yields the LHS.
This argument can be generalized to obtain the identity
$$
\sum_{i=0}^n \binom{i}{k}=\binom{n+1}{k+1}
$$
by classifying $k+1$-element subsets of $\{0,1,\dotsc, n\}$ based on their largest element.
