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JEE Main 2025 · MathsNumerical answerNumerical valueHardCalculation

JEE Main 3 April 2025, Shift 2, Maths Q21

Question 21 of 75 in this shift, Maths question 21 of 25, Section B.

Let

(1 + x + x^2)^{10} = a_0 + a_1 x + a_2 x^2 + \ldots + a_{20} x^{20}

. If

(a_1 + a_3 + a_5 + \ldots + a_{19}) - 11a_2 = 121k

, then

k

is equal to ____________.

Official answer

239

NTA final key.

Chapter: Binomial Theorem
Topic: Binomial Coefficients
Idea tested: Sum of Even and Odd Coefficients

Same idea in other shifts

Asked 2× in all

18 Mar 2021, Shift 1 · Q66Let $(1+x+2x^2)^{20} = a_0 + a_1x + a_2x^2 + \ldots + a_{40}x^{40}$ . Then, $a_1 + a_3 + a_5 + \ldots + a_{37}$ is equal to :HardSingle correct

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.