<s>[INST] 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?
 [/INST] $$\begin{align}
&1+(1+2)+(1+2+3)+\cdots+(1+2+3+\cdots+n)\\
&=n\cdot 1+(n-2)\cdot 2+(n-3)\cdot 3+\cdots +1\cdot n\\
&=\sum_{r=1}^n(n+1-r)r\\
&=\sum_{r=1}^n {n+1-r\choose 1}{r\choose 1}\\
&={n+2\choose 1+2}\\
&={n+2\choose 3}\\
&=\frac16 n(n+1)(n+2)
\end{align}$$
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