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4PM1/02 Jan 2022, Question 6: Sigma Notation

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 2 · Calculator · 13 marks

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The question, in short

Arithmetic series with Sₙ = n(3 + 2n); evaluate S₂₀, find constants A and B in the summation form, then solve Tₙ = Sₙ + 252 for a series with first term 7 and common difference 4.

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

Part (a) · 2 marks

Substitute n = 20 into Sₙ = n(3 + 2n) and calculate the value of S₂₀.

  1. 1S₂₀ = 20(3 + 2 × 20)
  2. 2= 20 × 43 = 860

Answer860

Part (b) · 6 marks

Given Sₙ equals the sum from r = 1 to n of (Ar + B), find the constants A and B.

  1. 1S₁ = 1(3 + 2) = 5, so the first term is 5
  2. 2S₂ = 2(3 + 4) = 14, so the second term is 14 − 5 = 9
  3. 3d = 9 − 5 = 4
  4. 4nth term = 5 + (n − 1) × 4 = 4n + 1
  5. 5So Sₙ = ∑ (4r + 1): A = 4, B = 1

AnswerA = 4, B = 1

Part (c) · 5 marks

For the series with first term 7 and common difference 4, use algebra to find n where Tₙ = Sₙ + 252.

  1. 1Tₙ = n/2 (2 × 7 + (n − 1) × 4) = n(5 + 2n)
  2. 2n(5 + 2n) = n(3 + 2n) + 252
  3. 35n + 2n² = 3n + 2n² + 252
  4. 42n = 252
  5. 5n = 126

Answern = 126

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