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Saturday, 31 October

JEE Main 2024 · MathsMultiple choiceSingle correctHardCalculation

JEE Main 4 April 2024, Shift 2, Maths Q5

Question 5 of 90 in this shift, Maths question 5 of 30, Section A.

The value of 1×22+2×32+…+100×(101)212×2+22×3+…+1002×101\frac{1\times 2^2+2\times 3^2+\ldots+100\times(101)^2}{1^2\times 2+2^2\times 3+\ldots+100^2\times 101} is
  1. (1)3130\frac{31}{30}
  2. (2)305301\frac{305}{301}Official answer
  3. (3)3231\frac{32}{31}
  4. (4)306305\frac{306}{305}

Official answer

Option 2

NTA final key.

Topic
No close match in our taxonomy
Idea tested
No close match in our taxonomy

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.