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

The sum k=1201+2+3++k is ___________.

Solution

    k=120k(k+1)2
=12k=120k2+k
=122021416+20212
=12420×416+20×212
=122870+210
=1540

Asked in: JEE Main 2020 (08 Jan Shift 1)

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