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:
$$\frac {1}{8} n (n + 1) (n^2 + n +2) $$
 A: The $k$th term is composed of $k$ integers, so the first $n$ terms, cumulatively, will be the composed of $\displaystyle F(n) = \sum_{k=1}^{n} k = \dfrac{n(n+1)}{2}$ consecutive integers overall. 
And to actually sum up those consecutive integers, we apply the same idea again.
$$S = F(F(n)) = F\left(\frac{n(n+1)}{2}\right) = \sum_{k=1}^{\frac{n(n+1)}{2}} k= \dfrac{\dfrac{n(n+1)}{2}\left(\dfrac{n(n+1)}{2}+1\right)}{2} = \frac{n (n^3 + 2 n^2 + 3 n + 2)}{8}$$
Edit: All of this relies on the fact that the sum of the first $n$ integers can be expressed in the closed form $F(n) = \dfrac{n(n+1)}{2}$ in general. 
So for this problem, we're first determining how many consecutive integers show up in the first $n$ terms, which is $F(n)$. But now we need to sum the first $F(n)$ terms, so it's like computing $F(F(n))$.
A: 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
$$S_n = T_{T_n} \\ = \frac{\frac{n(n+1)}{2}\cdot\left(\frac{n(n+1)}{2}+1\right)}{2} \\ = \frac{n(n+1)(n^2+n+2)}{8}$$
A: Hint... You have probably noticed already that the last term in each bracket is the $n$th triangular number $\frac n2(n+1)$ 
Therefore you need to evaluate the sum of the first $\frac n2(n+1)$ integers.
A: 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
$$\sum_{k=1}^n\frac{k}{2}(k^2+1)=\frac{1}{2}\sum_{k=1}^n(k^3+k)=\frac{n^2(n+1)^2}{8}+\frac{n(n+1)}{4}=\frac{n(n+1)(n^2+n+2)}{8}.$$
A: 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 $1+2+3+4+5=15$.
A: 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:
$$S_{\frac{n(n+1)}{2}}=\frac{1+\frac{n(n+1)}{2}}{2}\cdot \frac{n(n+1)}{2}=\frac{n(n+1)(n^2+n+2)}{8}$$ 
