A counting argument involving triangle numbers.

Question

Let $T_n$ be the $n$th triangle number.

Use a counting argument to prove:

$$nT_1+(n-1)T_2+...+T_n={{n+3}\choose{4}}$$

Edit: to clarify I have already proven this using standard summation formulae, I am just wondering if there is a super slick way to prove this using the "double counting" combinatorics technique. I'm a novice at this and have only seen it used a couple of times.

Answer 1

Hint: The right-hand side counts the number of four-element subsets of the $n + 3$ element set $$\{-1, 0, 1, 2, 3, \ldots, n, n + 1\}$$ The left-hand side counts the number of such subsets whose second-largest element is $k$. To make the argument work, you will need to show that $$T_k = \binom{k + 1}{2}$$

Answer 2

Alternatively (algebraic proof): $$\sum_{i=1}^n\sum_{j=1}^i T_j=\sum_{i=1}^n {i+2\choose 3}={n+3\choose 4}.$$