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?

Hint: use also that
$$
1^2 + 2^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}6
$$
$$
1 + (1+2) + \dots + (1 +2+\dots +n) =
\frac{1(1+1)}2 + \frac{2(2+1)}2 + \dots + \frac{n(n+1)}2 
\\=\frac 12 \left[
(1^2 + 1) + (2^2 + 2 ) + \dots + (n^2 + n)
\right]
\\=\frac 12 \left[
(1^2 + 2^2 + \dots + n^2) + (1 + 2 + \dots + n)
\right]
$$
