The sum ∑ k = 1 20 1 + 2 + 3 + … + k is ___________.

∑ k = 1 20 k ( k + 1 ) 2 = 1 2 ∑ k = 1 20 k 2 + k = 1 2 20 21 41 6 + 20 21 2 = 1 2 420 × 41 6 + 20 × 21 2 = 1 2 2870 + 210 = 1540

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Mathematics - Sequence & Series Question with Solution | TestHub

MathematicsSequence & SeriesSpecial sequences/seriesMedium2 minPYQ_2020

MathematicsMediumstatement

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

Answer:

1540

Solution:

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

Stream:JEESubject:MathematicsTopic:Sequence & SeriesSubtopic:Special sequences/series

⏱ 2mℹ️ Source: PYQ_2020

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