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

The first element in all sum is $\frac{n(n-1)}{2}+1$.
Thus, the sum of all elements in $n$-th sum is $$\frac{n^2(n-1)}{2}+1+2+...+n=\frac{n^2(n-1)}{2}+\frac{n(n+1)}{2}=\frac{n}{2}(n^2+1).$$
Thus the given sum is
$$\sum_{k=1}^n\frac{k}{2}(k^2+1)=\frac{1}{2}\sum_{k=1}^n(k^3+k)=\frac{n^2(n+1)^2}{8}+\frac{n(n+1)}{4}=\frac{n(n+1)(n^2+n+2)}{8}.$$
