The sum
Question:

The sum $\sum_{k=1}^{20}(1+2+3+\ldots+k)$ is________.

Solution:

Given series can be written as

$\sum_{k=1}^{20} \frac{k(k+1)}{2}=\frac{1}{2} \sum_{k=1}^{20}\left(k^{2}+k\right)$

$=\frac{1}{2}\left[\frac{20(21)(41)}{6}+\frac{20(21)}{2}\right]$

$=\frac{1}{2}\left[\frac{420 \times 41}{6}+\frac{20 \times 21}{2}\right]=\frac{1}{2}[2870+210]=1540$

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