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:
$$\frac {1}{8} n (n + 1) (n^2 + n +2) $$

Well, $(1)+(2+3)$ is the same as $1+2+3$, so if you have $n$ terms, then you can find the sum by $$\sum_{i=1}^n i = \frac{(n)(n+1)}{2}$$
Let us try for example, 5:
$\frac{(5)(6)}{2}$
And you get 15, then $1+2+3+4+5=15$.
