Let nth term of the series $ 1+(1+2)+(1+2+3)+(1+2+3+4)...(1+2+3+..n)$ be denoted as $a_{n}$
$a_{n} = \sum_1^n i$ $=\frac{n(n+1)}{2} = \frac{n^2 + n}{2}$
Sum of n-terms of series $S_{n}= \sum_1^na_{n} = \sum_1^n \frac{n^2+n}{2} = \frac{1}{2}*\frac{n(n+1)(2n+1)}{6}+ \frac{1}{2}*\frac{n(n+1)}{2}$
Further solving this, we get : $S_{n} = \frac{n(n+1)(2n+4)}{12}$
