Further Mathematics · Pearson Edexcel 9FM0 · Option D2 · Topic 6

Recurrence relations

Recurrence relations and their closed-form solutions.

  • 2 sub-topics
  • Taught in Year 13
  • 29 practice questions
  • 2 animated lessons
Practice questions (29) Lessons Worksheets All Edexcel Further Mathematics topics

What the specification says

Every sub-topic Pearson Edexcel can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

  1. 6.1Recurrence relations: modellingY13
  2. 6.2Solving recurrence relationsY13

Try 2 questions

Medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 29 questions in this topic, free, no account needed.

Recurrence relations

Not needed for Year 12Medium

Modelling with recurrence relations · Year 13 · 4 marks

A loan of £8000 is taken out. At the end of each month, interest of 1.5% of the amount owed is added, and then a repayment of £400 is made. Let unu_n be the amount owed after nn months.
  1. (a)
    Write down a recurrence relation for un+1u_{n + 1}, together with u0u_0.
    (2)
  2. (b)
    Find the amount owed after two months.
    (2)

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Part (a)2 marks

  1. Add the interest, then subtract the repayment.un+1=1.015un−400u_{n + 1} = 1.015u_n - 400
  2. u0=8000u_0 = 8000

Part (b)2 marks

  1. u1=7720.00u_1 = 7720.00
  2. u2=7435.80u_2 = 7435.80

Total4

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Recurrence relations

Not needed for Year 12Hard

Solving a first-order recurrence relation · Year 13 · 6 marks

un+1=1.02un−200u_{n + 1} = 1.02u_n - 200, u0=5000u_0 = 5000, models the amount owed on a loan after nn months.
  1. (a)
    Solve the recurrence relation.
    (4)
  2. (b)
    Find after how many months the loan is paid off.
    (2)

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Part (a)4 marks

  1. Try a constant.particular solution: u\text{particular solution: } u=1.02u−200{}= 1.02u - 200⇒u{}\Rightarrow u=10000{}= 10000
  2. General solution = complementary function + particular solution.un=A(1.02)n+10000u_n = A(1.02)^n + 10000
  3. Use u₀.un=10000−5000(1.02)nu_n = 10000 - 5000(1.02)^n

Part (b)2 marks

  1. 5000(1.02)n≥100005000(1.02)^n \geq 10000⇒n≥35.00{}\Rightarrow n \geq 35.00
  2. n=36 monthsn = 36\text{ months}

Total6

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