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JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Sequences and Series

The sum

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

is

Answer: 1540

Solution

The inner sum is

1 + 2 + \dots + k = \frac{k ( k + 1 )}{2}

, so the required total is

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

. Using

\sum_{k = 1}^{20} k^{2} = \frac{20 \cdot 21 \cdot 41}{6} = 2870

and

\sum_{k = 1}^{20} k = \frac{20 \cdot 21}{2} = 210

, the total is

\frac{1}{2} (2870 + 210) = \frac{3080}{2} = 1540

.

JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Integer answer

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