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# Engineering

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

Answers (1)
K Kuldeep Maurya

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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