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WMA12/01 Jun 2025, Question 1: The Binomial Expansion

Pearson Edexcel International AS Level Mathematics · Pure Mathematics 2 (WMA12) · Calculator · 6 marks

The question, in short

Expand (1 − 4x)⁷ up to the x³ term in simplest form, then use that expansion to determine the constant k given the x² coefficient of (5 + kx)(1 − 4x)⁷ is 1316.

Summarised in our own words. Use the official paper for the exact wording and any diagrams.

Part (a) · 4 marks

Write the first four terms of the binomial expansion of (1 − 4x)⁷ in ascending powers of x, simplifying each term.

  1. 1(1 − 4x)⁷ = 1 + ⁷C₁(−4x) + ⁷C₂(−4x)² + ⁷C₃(−4x)³ + …
  2. 2⁷C₁(−4x) = 7 × (−4x) = −28x
  3. 3⁷C₂(−4x)² = 21 × 16x² = 336x²
  4. 4⁷C₃(−4x)³ = 35 × (−64x³) = −2240x³
  5. 5So (1 − 4x)⁷ = 1 − 28x + 336x² − 2240x³ + …

Answer1 − 28x + 336x² − 2240x³

Part (b) · 2 marks

Using part (a), find k so that the x² coefficient in (5 + kx)(1 − 4x)⁷ equals 1316.

  1. 1x² terms in (5 + kx)(1 − 28x + 336x² + …): 5 × 336x² and kx × (−28x)
  2. 2Coefficient of x²: 1680 − 28k = 1316
  3. 328k = 364, so k = 13

Answerk = 13

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