$\text{ The }sum\sum _{k=1}^{20}(1+2+3+\dots ..+\mathrm{k})\text{ is }$

$\begin{array}{l}=\sum _{k=1}^{20}\frac{k(k+1)}{2}\\ =\frac{1}{2}\sum _{k=1}^{20}\left(k^2+k\right)\mid \end{array}$

$\begin{array}{l}=\frac{1}{2}\left[\frac{20\left(21\right)\left(41\right)}{6}+\frac{20\left(21\right)}{2}\right]\\ =\frac{1}{2}\left[\frac{420\times 41}{6}+\frac{20\times 21}{2}\right]\\ =\frac{1}{2}[2870+210]\\ =1540\end{array}$