### O(1) solution for: 1+(1+2)+(1+2+3)+...+(1+2+3+...+n)

We can solve this in $$O(n)$$ time if we apply the closed formula for sum of $$N$$ numbers. We need to traverse up till $$N$$ and for each iteration calculate the sum of $$N$$ numbers. But do you know what's even better? You can do it in $$O(1)$$ time. You just need to play with the same closed sum formula.

Let the $$N^{th} term$$ of the series be $$S_{N}$$,
$S_{N} = \sum_{i=1}^{N} i = \dfrac{N(N+1)}{2} = \dfrac{N^2+N}{2}\\ \sum_{i=1}^{N} S_{N} = \sum_{i=1}^{N} \dfrac{N^2+N}{2}\\ = \dfrac{1}{2} \sum_{i=1}^{N} N^2 + \sum_{i=1}^{N} N\\ = \dfrac{1}{2} \dfrac{N(N+1)(2N+1)}{6} + \dfrac{1}{2} \dfrac{N(N+1)}{2}\\ = \dfrac{1}{2}N(N+1) [\dfrac{2N+1}{6} + \dfrac{1}{2}]\\ = \dfrac{N(N+1)(2N+4)}{12}$
$\therefore S_{N} = \dfrac{N(N+1)(2N+4)}{12}$

Here is the code in C++: