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JEE Main 2021 · MathsMultiple choiceSingle correctHardMulti-step

JEE Main 18 March 2021, Shift 1, Maths Q66

Question 66 of 90 in this shift, Maths question 6 of 30, Section A.

Let

(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 :

(1) $2^{19}(2^{20}+21)$
(2) $2^{20}(2^{20}+21)$
(3) $2^{19}(2^{20}-21)$ Official answer
(4) $2^{20}(2^{20}-21)$

Official answer

Option 3

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

3 Apr 2025, Shift 2 · Q21Let $(1 + x + x^2)^{10} = a_0 + a_1 x + a_2 x^2 + \ldots + a_{20} x^{20}$ . If $(a_1 + a_3 + a_5 + \ldots + a_{19}) - 11a_2 = 121k$ , then…HardNumerical value

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.