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4PM1/02 Jun 2024, Question 7: Binomial Series for Any Rational Power

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 2 · Calculator · 7 marks

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The question, in short

Expand a binomial with a negative fractional power up to x⁶, then use it to build a series for a rational function and solve for k from a coefficient condition.

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

Part (a) · 3 marks

Write (1 + 2x²)^(−3/4) as a series in ascending powers of x through x⁶, giving each coefficient as a fraction in lowest terms.

  1. 1(1 + 2x²)^(−3/4) = 1 + (−3/4)(2x²) + (−3/4)(−7/4)(2x²)²/2! + (−3/4)(−7/4)(−11/4)(2x²)³/3!
  2. 2= 1 − (3/2)x² + (21/8)x⁴ − (77/16)x⁶

Answer1 − (3/2)x² + (21/8)x⁴ − (77/16)x⁶

Part (b) · 2 marks

Expand f(x) = (2 + kx)/(1 + 2x²)^(3/4) as a series up to x⁵, expressing coefficients in terms of k.

  1. 1f(x) = (2 + kx)(1 − (3/2)x² + (21/8)x⁴ − …)
  2. 2= 2 + kx − 3x² − (3k/2)x³ + (21/4)x⁴ + (21k/8)x⁵

Answer2 + kx − 3x² − (3k/2)x³ + (21/4)x⁴ + (21k/8)x⁵

Part (c) · 2 marks

Given that the x⁵ coefficient equals fourteen times the x² coefficient, determine k.

  1. 1Coefficient of x⁵ = 21k/8, coefficient of x² = −3
  2. 221k/8 = 14 × (−3) = −42
  3. 3k = −42 × 8/21 = −16

Answerk = −16

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