How to find the sum of the series $(1) + (2 + 3) +(4 + 5 + 6) + \cdots$ to $n$ terms? How to find the sum of this series?
$$(1) + (2 + 3) +(4 + 5 + 6) + \cdots \text{ to }\, n \, \text{ terms }$$
Answer is given as:
$$\frac {1}{8} n (n + 1) (n^2 + n +2) $$

Let $T_n$ be the $n$th triangle number, which are defined by
$$T_n := \sum_{k=1}^n k = 1 + 2 +...+n.$$
Then we can see that the sum you are looking for (which wee will call $S_n$) equates to
$$S_n = T_{T_n}$$
since $S_n$ sums the first $T_n$ natural numbers.
Thus, using the fact that $T_n = \frac{n(n+1)}{2}$, we have
$$S_n = T_{T_n} \\ = \frac{\frac{n(n+1)}{2}\cdot\left(\frac{n(n+1)}{2}+1\right)}{2} \\ = \frac{n(n+1)(n^2+n+2)}{8}$$
