Question

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

Answer 1

Simplify the given function using the formula of Arithmetic Progression :

$\begin{array}{rcl}& & \sum _{k=1}^{20}(1+2+3+.......+k)\\ & =& \sum _{k=1}^{20}\left(\frac{k(k+1)}{2}\right)\\ & =& \frac{1}{2}\sum _{k=1}^{20}(k^2+k)\\ & =& \frac{1}{2}[\frac{20(20+1)(20\times 2+1)}{6}+\frac{20(1+20)}{2}]\\ & =& \frac{1}{2}[\frac{20\times 21\times 41}{6}+\frac{20\times 21}{2}]\\ & =& 1540\end{array}$

Hence , answer is $=1540$.