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 4× in all

25 Jun 2022, Shift 2 · Q24If the sum of the co-efficients of all the positive even powers of $x$ in the binomial expansion of $\left(2x^3 + \frac{3}{x}\right)^{10}$ …HardNumerical value
4 Aug 2021, Shift 1 · Q65If $(1+x+x^2)^4=a_0+a_1x+a_2x^2+\ldots\ldots+a_8x^8$ , then the value of $a_0+4a_2+16a_4+\ldots.+2^8a_8$ is equal toMediumSingle correct
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 108 of the 174 JEE Main shifts held from 2017 to 2026.