The sum∑$$k=120(1+2+3+$$…..$$+k)$$ is

$$=$$∑$$k=120k(k+1)2=12$$∑$$k=120k2+k$$∣$$=1220(21)(41)6+20(21)2=12420$$×$$416+20$$×$$212=12$$[$$2870+210$$]$$=1540$$

Sigma operation series

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

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Solution

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

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Sigma operation series

Remember concepts with our Masterclasses.

The sum∑k=120(1+2+3+…..+k) is

=∑k=120k(k+1)2=12∑k=120k2+k∣=1220(21)(41)6+20(21)2=12420×416+20×212=12[2870+210]=1540

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