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WMA12/01 Oct 2023, Question 10: The Binomial Expansion

Pearson Edexcel International AS Level Mathematics · Pure Mathematics 2 (WMA12) · Calculator · 12 marks

The question, in short

Expand a binomial, use three of its terms as a geometric series, and solve a trigonometric geometric series with sum to infinity 8/5.

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

Part (ia) · 3 marks

Write the 2nd, 3rd and 5th terms of (3 + 2x)⁶ in ascending powers of x.

  1. 12nd term: ⁶C₁ × 3⁵ × (2x) = 6 × 243 × 2x = 2916x
  2. 23rd term: ⁶C₂ × 3⁴ × (2x)² = 15 × 81 × 4x² = 4860x²
  3. 35th term: ⁶C₄ × 3² × (2x)⁴ = 15 × 9 × 16x⁴ = 2160x⁴

Answer2nd term 2916x, 3rd term 4860x², 5th term 2160x⁴

Part (ib) · 3 marks

Find the x for which those three terms are consecutive terms of a geometric series.

  1. 1Consecutive GP terms have a common ratio: 4860x²/2916x = 2160x⁴/4860x²
  2. 2So (4860x²)² = 2916x × 2160x⁴
  3. 323 619 600x⁴ = 6 298 560x⁵
  4. 4x = 23 619 600/6 298 560 = 15/4 (x ≠ 0)

Answerx = 15/4

Part (iia) · 3 marks

Use first term sin²θ, ratio 2 cos θ and sum to infinity 8/5 to show 5 cos²θ − 16 cos θ + 3 = 0.

  1. 1S∞ = a/(1 − r): 8/5 = sin²θ/(1 − 2cos θ)
  2. 2Use sin²θ = 1 − cos²θ: 8/5 = (1 − cos²θ)/(1 − 2cos θ)
  3. 3Cross-multiply: 8 − 16cos θ = 5 − 5cos²θ
  4. 4So 5cos²θ − 16cos θ + 3 = 0

AnswerShown: 8/5 = sin²θ/(1 − 2cos θ) = (1 − cos²θ)/(1 − 2cos θ) ⇒ 8 − 16cos θ = 5 − 5cos²θ ⇒ 5cos²θ − 16cos θ + 3 = 0

Part (iib) · 3 marks

Hence find the exact 2nd term of that series.

  1. 1(5cos θ − 1)(cos θ − 3) = 0, so cos θ = 1/5 (cos θ = 3 is impossible)
  2. 2sin²θ = 1 − 1/25 = 24/25, and r = 2 × 1/5 = 2/5
  3. 32nd term = ar = (24/25) × (2/5) = 48/125

Answer48/125

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