Combinatoric meaning to $1+2+\dots+n=\frac{n(n+1)}{2}= {n+1 \choose 2}$

How many pairs can we obtain with $n+1$ objects?

• first and second, first and third, $\,\ldots= n$ pairs
• second and third, second and fourth, $\,\ldots= n-1$ pairs
• and so on $\,\ldots$

$1+2+3+4+\cdots+n\\ =\binom{1}{1}+\binom{2}{1}+\binom{3}1+\binom{4}1+\cdots+\binom{n}1\\ =\binom{2}{2}+\binom{2}1+\binom{3}1+\binom{4}1+\cdots+\binom{n}1$

and use $\binom{a}{b+1}+\binom{a}{b}=\binom{a+1}{b+1}$.

This may not be what you are looking for, but I think it's a little cool.