The coefficient of x²⁰¹² in the expansion of (1-x)²⁰⁰⁸(1+x+x²)²⁰⁰⁷ is equal to _________.

Solution

Maths sir · 4 steps

Pair (1-x) with (1+x+x²) to get 1-x³, and only powers of 3 are left to deal with.

1. Combine the factors. Since (1-x)(1+x+x²) = 1-x³, take 2007 factors of (1-x) and pair each with one (1+x+x²). One factor of (1-x) is left over.

   (1-x)²⁰⁰⁸(1+x+x²)²⁰⁰⁷ = (1-x)(1-x³)²⁰⁰⁷

2. Which powers appear. (1-x³)²⁰⁰⁷ contains only powers x³ʳ. Multiplying by (1-x) gives terms x³ʳ and x³ʳ⁺¹ only. So every power in the product leaves remainder 0 or 1 when divided by 3.
3. Check 2012. 2012 = 3 × 670 + 2, so its remainder is 2. To get x²⁰¹² you would need x²⁰¹² or x²⁰¹¹ from (1-x³)²⁰⁰⁷, and neither 2012 nor 2011 is a multiple of 3.
4. So. No term x²⁰¹² appears in the expansion, so its coefficient is 0.

The idea to keep. When you see 1+x+x² next to 1-x, multiply them into 1-x³ first. Then only powers of a fixed pattern survive, and a coefficient can be zero just by divisibility.

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