JEE Main 2021 · 20 July · morning · Single correct · +4 / −1

The coefficient of x²⁵⁶ in the expansion of
(1 - x)¹⁰¹ (x² + x + 1)¹⁰⁰ is :

  1. (a) ¹⁰⁰C₁₆
  2. (b) ¹⁰⁰C₁₅
  3. (c) - ¹⁰⁰C₁₆
  4. (d) - ¹⁰⁰C₁₅
Answer and solution

Answer (b) ¹⁰⁰C₁₅

Solution

Maths sir · 4 steps

Pair one (1-x) with (1+x+x²) each time: their product is 1-x³, and the whole thing becomes easy.

  1. Use the identity. (1-x)(1+x+x²) = 1-x³. Split (1-x)¹⁰¹ as (1-x)×(1-x)¹⁰⁰ and pair the 100 factors with (x²+x+1)¹⁰⁰.

    (1-x)¹⁰¹(x²+x+1)¹⁰⁰ = (1-x)(1-x³)¹⁰⁰ = (1-x)Σₖ₌₀¹⁰⁰ (-1)ᵏ ¹⁰⁰Cₖ x³ᵏ

  2. Find the terms. Every power in (1-x³)¹⁰⁰ is a multiple of 3. The 1 would need x²⁵⁶, but 256 is not a multiple of 3, so it gives nothing. The -x needs x²⁵⁵, that is 3k = 255, so k = 85.
  3. Track the signs. The coefficient is (-1)×(-1)⁸⁵ ¹⁰⁰C₈₅ = (-1)×(-1)×¹⁰⁰C₈₅ = ¹⁰⁰C₈₅. Since ⁿCᵣ = ⁿCₙ₋ᵣ, this equals ¹⁰⁰C₁₅.
  4. So. The coefficient of x²⁵⁶ is ¹⁰⁰C₁₅, which is option (b).

The idea to keep. When (1-x) and (1+x+x²) appear together, combine them into 1-x³ first. Then only powers that are multiples of 3 survive, and the extra factor shifts the power you need.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Binomial Theorem

Asked about 200 times in JEE Main since 2020. This question is from 2021.