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WMA12/01 Oct 2020, Question 1: The Binomial Expansion

Pearson Edexcel International AS Level Mathematics · Pure Mathematics 2 (WMA12) · Calculator · 7 marks

The question, in short

Expand (2 − x/4)¹⁰ as a binomial series up to the x³ term, then use that result to locate the term with no x in (3 − 1/x)² times the same bracket.

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

Part (a) · 4 marks

Write out the first four terms of the expansion of (2 − x/4)¹⁰ in increasing powers of x, simplifying each coefficient fully.

  1. 1(2 − x/4)¹⁰ = 2¹⁰ + ¹⁰C₁·2⁹·(−x/4) + ¹⁰C₂·2⁸·(−x/4)² + ¹⁰C₃·2⁷·(−x/4)³ + …
  2. 22¹⁰ = 1024 and 10 × 512 × (−x/4) = −1280x
  3. 345 × 256 × x²/16 = 720x²
  4. 4120 × 128 × (−x³/64) = −240x³
  5. 5So (2 − x/4)¹⁰ = 1024 − 1280x + 720x² − 240x³ + …

Answer1024 − 1280x + 720x² − 240x³

Part (b) · 3 marks

Using your expansion from part a, determine the constant term in the product (3 − 1/x)²(2 − x/4)¹⁰.

  1. 1(3 − 1/x)² = 9 − 6/x + 1/x²
  2. 2Constant term comes from 9 × 1024, (−6/x) × (−1280x) and (1/x²) × 720x²
  3. 39 × 1024 + 7680 + 720
  4. 4= 9216 + 7680 + 720 = 17616

Answer17616

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Also searched as: WMA12 October 2020 · WMA12/01 · wma12 oct20 · IAS Maths Pure Mathematics 2 October 2020 · question 1.
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