1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) + ...

Solution:

Let $T_{n}$ be the $n$th term of the given series.

Thus, we have:

$T_{n}=1+2+3+4+5+\ldots+n=\frac{n(n+1)}{2}=\frac{n^{2}+n}{2}$

Now, let $S_{n}$ be the sum of $n$ terms of the given series.

Thus, we have:

$S_{n}=\sum_{k=1}^{n} T_{k}$

$\Rightarrow S_{n}=\sum_{k=1}^{n}\left(\frac{k^{2}+k}{2}\right)$

$\Rightarrow S_{n}=\frac{1}{2} \sum_{k=1}^{n}\left(k^{2}+k\right)$

$\Rightarrow S_{n}=\frac{1}{2}\left[\frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}\right]$

$\Rightarrow S_{n}=\frac{n(n+1)}{4}\left(\frac{2 n+1}{3}+1\right)$

$\Rightarrow S_{n}=\frac{n(n+1)}{4}\left(\frac{2 n+4}{3}\right)$

$\Rightarrow S_{n}=\frac{n(n+1)(2 n+4)}{12}$

$\Rightarrow S_{n}=\frac{n(n+1)(n+2)}{6}$