The sum $\sum_{k=1}^{20}(1+2+3+.....+k)$ is Option: 1 1540 Option: 2 1680 Option: 3 1260 Option: 4 1450

Sum of Common Series

Sum of the square of first n-natural numbers

$\\\mathrm{1^2+2^2+3^2+4^2+...........+n^2=\frac{n(n+1)(2n+1)}{6}}$

Now,

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

$=\frac{1}{2}\left(\frac{20(20+1)(2 \times 20+1)}{6}+\frac{20(20+1)}{2}\right)$

$=\frac{1}{2}((70 \times 41)+210)=1540$

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