O-Level Binomial expansions: worked solution

Question

Using (1 + 3x)^8 = 1 + 24x + 252x² + ..., deduce the first three terms of (2 + 6x + 3x²)^8.

The expansion supplied is for a bracket beginning with 1, so first make 2 + 6x + 3x² begin with 1 by taking out the factor 2: 2 + 6x + 3x² = 2(1 + 3x + (3/2)x²). Raise both sides to the power 8, remembering that the 2 outside is raised to the power 8 as well: (2 + 6x + 3x²)⁸ = 2⁸ (1 + 3x + (3/2)x²)⁸ = 256(1 + 3x + (3/2)x²)⁸. Now expand (1 + 3x + (3/2)x²)⁸ as far as x². Group the bracket as (1 + 3x) + (3/2)x² so that the given expansion of (1 + 3x)⁸ can be used. With u = 1 + 3x and v = (3/2)x², the binomial theorem gives (u + v)⁸ = u⁸ + 8u⁷v + 28u⁶v² + ... The term 28u⁶v² contains v² = (9/4)x⁴, so it only affects x⁴ and beyond and may be dropped. First term: u⁸ = (1 + 3x)⁸ = 1 + 24x + 252x² + ... (this is the result given). Second term: 8u⁷v = 8(1 + 3x)⁷ × (3/2)x² = 12x²(1 + 3x)⁷. Since this is already multiplied by x², only the leading 1 of (1 + 3x)⁷ can contribute at the x² level, so 8u⁷v = 12x² + ... Add the two contributions: (1 + 3x + (3/2)x²)⁸ = 1 + 24x + (252 + 12)x² + ... = 1 + 24x + 264x² + ... Finally multiply every term by the 256 taken out at the start: 256 × 1 = 256 256 × 24 = 6144 256 × 264 = 67584. So (2 + 6x + 3x²)⁸ = 256 + 6144x + 67584x² + ...