The coefficient of x⁴ in the expansion of
(1 + x + x² + x³)⁶ in powers of x, is ______.

Solution

Maths sir · 3 steps

Factor 1 + x + x² + x³ as (1+x)(1+x²), and the whole problem becomes two easy binomials.

1. Factor first. Group the terms: 1 + x + x² + x³ = (1+x) + x²(1+x) = (1+x)(1+x²). So the expression is (1+x)⁶(1+x²)⁶.

   (1+x+x²+x³)⁶ = (Σᵣ₌₀⁶ C(6, r) xʳ)(Σⱼ₌₀⁶ C(6, j) x²ʲ)

2. Pair the powers. To get x⁴ you need r + 2j = 4. The only choices are j = 0, r = 4, then j = 1, r = 2, and j = 2, r = 0.

   j (from (1+x²)⁶)	r (from (1+x)⁶)	Contribution
   0	4	C(6, 0)C(6, 4) = 1 × 15 = 15
   1	2	C(6, 1)C(6, 2) = 6 × 15 = 90
   2	0	C(6, 2)C(6, 0) = 15 × 1 = 15

3. So. The coefficient of x⁴ is 15 + 90 + 15 = 120.

The common slip: 126. Counting with stars and bars, C(9, 4), gives 126. That count allows one bracket to supply x⁴, which 1 + x + x² + x³ cannot do, so the 6 such cases must be removed.

The idea to keep. Before expanding a multinomial, try to factor the base into binomials. Then list every pair of powers that adds up to the power you need, and add their products.

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