Further Mathematics · OCR H245 · Option A · Topic 8.01

Sequences and series

Recurrence relations, properties of sequences, Fibonacci and related numbers, solving recurrence systems, modelling.

  • 5 sub-topics
  • Taught in Year 13
  • 76 practice questions
  • 5 animated lessons
Practice questions (76) Lessons Worksheets coming soon All OCR Further Mathematics topics

What the specification says

Every sub-topic OCR 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. 8.01a-bRecurrence relationsY13
  2. 8.01c-dProperties of sequencesY13
  3. 8.01eFibonacci and related numbersY13
  4. 8.01f-gSolving recurrence systemsY13
  5. 8.01h-iModellingY13

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 76 questions in this topic, free, no account needed.

Further sequences and series

Not needed for Year 12Easy

Forming a recurrence · Year 13 · 1 mark

A savings account earns 2% per year and £100 is added each year.
  1. Write a recurrence for the balance uₙ.
    (1)
Show worked solutionHide worked solution
  1. un+1=1.02un+100u_{n+1} = 1.02u_n + 100

Total1

  • M
  • A
  • B
How did it go?

Report a problem with this question

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)

Check your answers before you look at the solution

Show worked solutionHide worked solution

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

  • M
  • A
  • B
How did it go?

Report a problem with this question

Further sequences and series

Not needed for Year 12Hard

Forming a recurrence relation · Year 13 · 6 marks

A code is a string of letters from the set {A,B,C,D}\{A, B, C, D\}. A code is valid if it never has two A's next to each other. Let unu_n be the number of valid codes of length nn.
  1. (a)
    Explain why un=3un−1+3un−2u_n = 3u_{n - 1} + 3u_{n - 2} for n≥3n \geq 3.
    (3)
  2. (b)
    Find u1u_1 and u2u_2, and hence find u6u_6.
    (3)

Check your answer before you look at the solution

Show worked solutionHide worked solution

Part (a)3 marks

  1. Split by the last letter.A valid code ending in a letter other than AA: 3 choices for the last letter, after any valid code of length n−1n - 1.
  2. A valid code ending in A: the letter before must not be A (3 choices), after any valid code of length n − 2.
  3. un=3un−1+3un−2u_n = 3u_{n - 1} + 3u_{n - 2}

Part (b)3 marks

  1. Only AA is invalid at length 2.u1=4, u2=16−1=15u_1 = 4, \ u_2 = 16 - 1 = 15
  2. u3=57, u4=216, u5=819u_3 = 57, \ u_4 = 216, \ u_5 = 819
  3. u6=3105u_6 = 3105

Total6

  • M
  • A
  • B
How did it go?

Report a problem with this question