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4PM1/01 Jan 2023, Question 6: Binomial Series for Any Rational Power

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 1 · Calculator · 7 marks

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The question, in short

Rewrite (8 + 3x)^(1/3) as p(1 + qx)^(1/3), expand up to the x² term with exact fraction coefficients, then use the expansion to show an approximation for ∛9.

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

Part (a) · 2 marks

Find the constants p and q so that (8 + 3x)^(1/3) equals p(1 + qx)^(1/3).

  1. 1(8 + 3x)^(1/3) = 8^(1/3)(1 + 3x/8)^(1/3)
  2. 2= 2(1 + (3/8)x)^(1/3)

Answerp = 2, q = 3/8

Part (b) · 3 marks

Using your p and q, expand (8 + 3x)^(1/3) in ascending powers of x through x², giving each coefficient as a lowest-terms fraction.

  1. 1(1 + (3/8)x)^(1/3) = 1 + (1/3)(3x/8) + ((1/3)(−2/3)/2)(3x/8)² + …
  2. 2= 1 + x/8 − x²/64 + …
  3. 3Multiply by 2: 2 + x/4 − x²/32

Answer2 + (1/4)x − (1/32)x²

Part (c) · 2 marks

Choose a suitable x in the part (b) expansion and show that ∛9 is approximately 599/288.

  1. 18 + 3x = 9 when x = 1/3
  2. 2∛9 ≈ 2 + (1/4)(1/3) − (1/32)(1/9)
  3. 3= 2 + 1/12 − 1/288 = (576 + 24 − 1)/288 = 599/288

Answer∛9 ≈ 599/288 shown

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