JEE Main 2019MathematicsSequences and SeriesMediumMCQ

JEE Main 2019 — Sequences and Series Question with Solution

JEE Main 2019 (12 Jan Shift 1)

Question

Let Sk=1 + 2 + 3+…+kk. If S12+S22+…+S102=512A, then A is equal to :

Choose an option

Show full solutionCorrect option: B
Correct answer
B303

Step-by-step explanation

Sk=1+2+3+…+kk=kk+12k=k+12 .........i

∴ S12+S22+....+S102=1+122+2+122+...+10+122  (using equation i)

=222+322+...+1122

=12222+32+...+112

=1412+22+32+...+112-12

=1411×12×236-1 (using the identity ∑n2n=1n=n=nn+12n+16)

=14×505

Comparing with the given condition in question, we get

512A=5054

⇒A=505×35=303.

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About this question

This is a previous-year question from JEE Main 2019, covering the Sequences and Series chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.