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4PM1/02R Jun 2022, Question 9: Binomial Series for Any Rational Power

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 2R · Calculator · 9 marks

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The question, in short

Rewrite the fraction 3/(3 − x)³ in the required power form, expand it in ascending powers of x up to x³, then use the expansion to approximate 24/125 and find the percentage error.

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

Part (a) · 2 marks

Express 3/(3 − x)³ as a(1 − bx)^(−3), giving a and b as fractions in lowest terms.

  1. 1(3 − x)³ = 27(1 − x/3)³
  2. 23/(3 − x)³ = (3/27)(1 − x/3)^(−3) = (1/9)(1 − x/3)^(−3)
  3. 3a = 1/9, b = 1/3

Answera = 1/9, b = 1/3

Part (b) · 3 marks

Find the binomial expansion of 3/(3 − x)³ in ascending powers of x through the x³ term, with each coefficient as a lowest-terms fraction.

  1. 1(1 − x/3)^(−3) = 1 + (−3)(−x/3) + (−3)(−4)/2! (−x/3)² + (−3)(−4)(−5)/3! (−x/3)³
  2. 2= 1 + x + (2/3)x² + (10/27)x³
  3. 3Multiply by 1/9: 1/9 + (1/9)x + (2/27)x² + (10/243)x³

Answer1/9 + (1/9)x + (2/27)x² + (10/243)x³

Part (c) · 4 marks

Pick a suitable x, substitute into the part (b) expansion to estimate 24/125 to 5 decimal places, then calculate the percentage error to 2 decimal places against the true value.

  1. 124/125 = 3/(3 − x)³ gives (3 − x)³ = 125/8, so 3 − x = 5/2, x = 0.5
  2. 21/9 + (1/9)(0.5) + (2/27)(0.5)² + (10/243)(0.5)³ = 0.19033
  3. 3Actual value 24/125 = 0.192
  4. 4Percentage error = (0.192 − 0.19033)/0.192 × 100 = 0.87%

Answer(i) 0.19033 (ii) 0.87%

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