Simplify triangular sum of triangular numbers: $\sum_{i=1}^{n}(\frac12i(i+1))$

A good way is to simplify the summation as follows: $$\sum_{i=1}^n \frac{i(i+1)}{2}=\frac{1}{2}\left[\sum_{i=1}^n i^2 + \sum_{i=1}^n i \right]$$ and then use the standard results for $\sum_\limits{i=1}^n i^2$ and $\sum_\limits{i=1}^n i$.

In general we have:$$\sum_{i=r}^{n}\binom{i}{r}=\binom{n+1}{r+1}$$ This is a useful equality that can be proved with induction on $n$.

The induction step is (triangle of Pascal): $$\binom{n+1}{r+1}+\binom{n+1}{r}=\binom{n+2}{r+1}$$

You can apply this to find: $$\sum_{i=1}^n\frac{i(i+1)}2=\sum_{i=1}^{n}\binom{i+1}{2}=\sum_{i=2}^{n+1}\binom{i}{2}=\binom{n+2}{3}$$

It is possible to give a direct combinatorial proof that:

$$\sum_{i=1}^{n-2}T_i=\binom n3$$

which follows the logic of this answer on a related question. https://math.stackexchange.com/a/2478631/627744

Suppose $n$ people wish to take every possible 3-person selfie. By definition, there are $\binom n3$ ways to do this.

Alternatively, imagine the $n$ people are all standing in a line. First compose all the selfies in which person 1 participates. Person 1 follows as person 2 walks down the line taking a pic with each of the remaining $n-2$ people. Then person 2 steps aside so that person 1 can follow person 3, taking all $n-3$ 3-person pics that include persons 1 and 3, but not 2. Then person 3 steps aside so that person 1 can take the $n-4$ pics that include person 4, but not persons 2 or 3. This process continues so that there are $T_{n-2}$ selfies with person 1.

Person 1 now steps aside permanently and we compose all the selfies in which person 2 participates, but person 1 doesn't. Person 2 now plays the previous role of person 1, following each person down the row, taking $T_{n-3}$ selfies in total.

In this way, $\sum_{i=1}^{n-2}T_i$ counts the total number of 3-person selfies.