<s>[INST] 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:
$$\frac {1}{8} n (n + 1) (n^2 + n +2) $$
 [/INST] The last element in the $n$-th bracket is $\frac{n(n+1)}{2}$.
Thus, it is the arithmetic progression: 1, 2, 3, ...,$\frac{n(n+1)}{2}$, whose sum is:
$$S_{\frac{n(n+1)}{2}}=\frac{1+\frac{n(n+1)}{2}}{2}\cdot \frac{n(n+1)}{2}=\frac{n(n+1)(n^2+n+2)}{8}$$ 
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