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Friday, 9 October

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JEE Main 2024 · MathsMultiple choiceSingle correctMediumCalculation

JEE Main 6 April 2024, Shift 2, Maths Q9

Question 9 of 90 in this shift, Maths question 9 of 30, Section A.

\lim_{n\to\infty}\frac{(1^2-1)(n-1)+(2^2-2)(n-2)+\cdots+((n-1)^2-(n-1))\cdot 1}{(1^3+2^3+\cdots+n^3)-(1^2+2^2+\cdots+n^2)}

is equal to :

(1) $\frac{1}{2}$
(2) $\frac{1}{3}$ Official answer
(3) $\frac{2}{3}$
(4) $\frac{3}{4}$

Official answer

Option 2

NTA final key.

Chapter: Limits and Derivatives
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 42 of the 174 JEE Main shifts held from 2017 to 2026.