Calculate $1+(1+2)+(1+2+3)+\cdots+(1+2+3+\cdots+n)$ How can I calculate $1+(1+2)+(1+2+3)+\cdots+(1+2+3+\cdots+n)$? I know that $1+2+\cdots+n=\dfrac{n+1}{2}\dot\ n$. But what should I do next?

Sums we know:
$\sum^n_{i=1} i = 1+2+\cdots+n=\frac{n^2+n}{2}$
$\sum^n_{i=1} i^2 = 1^2 + 2^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}6$
Your sum is $$(1+2+3+ \cdots + n) + (1 + 2 + \cdots + (n-1)) + (1 + 2 + \cdots + (n-2)) + \cdots + (1)$$ $$= \sum^n_{k=1} \sum^k_{i=1} i$$ $$= \sum^n_{k=1} \frac{k^2+k}{2} = \frac 12 (\sum^n_{k=1} k^2 + \sum^n_{k=1} k)$$
NOTE: You can reorder the terms if the are a finite number of them.  So if you're going to be taking a limit as $n \to \infty$ don't do it this way.
