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WMA12/01 Jan 2021, Question 8: Sequences: Formulas and Recurrence

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

The question, in short

A recurrence gives each term from the previous one using aₙ₊₁ = 2(aₙ + 3)² − 7 with a₁ = p − 3; find a₂ in terms of p, then use the sum of the first three terms equalling p + 15 to find possible a₂ values.

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Part (a) · 1 mark

Substitute a₁ = p − 3 into the recurrence and simplify the resulting expression for a₂ in terms of p.

  1. 1a₂ = 2(a₁ + 3)² − 7 = 2(p − 3 + 3)² − 7
  2. 2a₂ = 2p² − 7

Answera₂ = 2p² − 7

Part (b) · 6 marks

Using the given sum of a₁, a₂ and a₃ equal to p + 15, form an equation and solve for the possible values of a₂.

  1. 1a₃ = 2(2p² − 7 + 3)² − 7 = 2(2p² − 4)² − 7 = 8p⁴ − 32p² + 25
  2. 2a₁ + a₂ + a₃ = (p − 3) + (2p² − 7) + (8p⁴ − 32p² + 25) = p + 15
  3. 38p⁴ − 30p² = 0, so 2p²(4p² − 15) = 0
  4. 4p² = 0 or p² = 15/4
  5. 5a₂ = 2(0) − 7 = −7 or a₂ = 2(15/4) − 7 = 1/2

Answera₂ = −7 or a₂ = 1/2

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