A Level Math Revision Free diagnostic
Themed maths · 23 November

Fibonacci Day maths for A Level

23 November is Fibonacci Day: written month first, 11/23 reads 1, 1, 2, 3. This ready-to-teach lesson has a 5-minute starter, a 35-minute main activity on recurrence relations, the ratio limit and the golden ratio in a regular pentagon, an extension that derives Binet’s formula, and full worked answers.

Level
A Level Maths (Year 12 and Year 13)
Time
40 minutes, plus a 10-minute extension
Topics
Recurrence relations and periodic sequences; Limits of sequences; Solving quadratics with surds; The cosine rule and exact values; Binet’s formula (extension)
Equipment
The starter is non-calculator. A calculator helps in tasks A and C.

Download student sheet (PDF)Answers (PDF)

Suggested timings

PartTimeWhat
Starter5 minQuick questions on the board
Main: task A12 minThe ratio limit
Main: task B11 minPeriodic and patterned sequences
Main: task C12 minThe golden ratio in a pentagon
Extension10 minFast finishers or homework

Starter (5 minutes)

No calculator. F1 = F2 = 1 and Fn+2 = Fn+1 + Fn.

  1. A sequence has u1 = 2, u2 = 1 and un+2 = un+1 + un. Find u7.
  2. Solve x2 − x − 1 = 0 exactly.
  3. Find Σr=18 Fr.
  4. φ = (1 + √5)/2. Rationalise the denominator to show that 1/φ = φ − 1.

Main activity (35 minutes)

Task A: The ratio limit (12 min)

Let rn = Fn+1 ÷ Fn.

  1. Find r1 to r6.
  2. Show that rn+1 = 1 + 1/rn, and use it to find r7 from r6.
  3. Assuming the sequence converges to L, show that L2 − L − 1 = 0. Find L to 3 decimal places and explain why you reject the other root.
  4. The terms lie alternately above and below L. Find r10 and how far it is from L, to 2 significant figures.

Task B: Periodic and patterned sequences (11 min)

Recurrence relations can repeat.

  1. A sequence has u1 = 2 and un+1 = 1/(1 − un). Find u2, u3 and u4. What is the period?
  2. Find Σr=130 ur for the sequence in part (a).
  3. The Fibonacci numbers follow the pattern odd, odd, even, odd, odd, even, … How many of F1, …, F100 are odd?

Task C: The golden ratio in a pentagon (12 min)

A regular pentagon has sides of length 1.

  1. Show that each interior angle is 108°.
  2. Use the cosine rule to find the length of a diagonal, to 4 significant figures.
  3. Explain why the diagonal is also 2 cos 36°, and given that cos 36° = (1 + √5)/4, show that the diagonal is exactly φ.

Extension (10 minutes)

Deriving Binet’s formula. Let φ = (1 + √5)/2 and ψ = (1 − √5)/2, the roots of x2 = x + 1.

  1. Show that un = Aφn + Bψn satisfies un+2 = un+1 + un for any constants A and B. Then use F1 = F2 = 1 to find A and B.
  2. Explain why Fn is the whole number nearest to φn/√5, and use this to find F25.

For teachers

Teacher notes and full worked answers
  • Fibonacci Day is 23 November because 11/23, written month first, reads 1, 1, 2, 3.
  • In task A (c), the assumption that the limit exists matters: ask the class for a recurrence where ‘solving for the limit’ gives a wrong answer because the sequence diverges.
  • Task C links the golden ratio to exact trigonometric values: cos 36° = φ/2.
  • The extension is a first look at second-order recurrences, which Further Maths students meet again.

Starter

  1. 18
    • 2, 1, 3, 4, 7, 11, 18
  2. x = (1 ± √5)/2
    • x = (1 ± √(1 + 4))/2
  3. 54
    • 1 + 1 + 2 + 3 + 5 + 8 + 13 + 21 = 54
  4. 1/φ = (√5 − 1)/2 = φ − 1
    • 1/φ = 2/(1 + √5) = 2(√5 − 1)/((√5 + 1)(√5 − 1)) = 2(√5 − 1)/4 = (√5 − 1)/2.
    • φ − 1 = (1 + √5)/2 − 1 = (√5 − 1)/2. They are equal.

Task A: The ratio limit

  1. 1, 2, 1.5, 1.667, 1.6, 1.625
    • 1/1, 2/1, 3/2, 5/3, 8/5, 13/8
  2. r7 = 21/13
    • Fn+2/Fn+1 = (Fn+1 + Fn)/Fn+1 = 1 + 1/rn.
    • r7 = 1 + 8/13 = 21/13
  3. L = (1 + √5)/2 = 1.618
    • In the limit L = 1 + 1/L, so L2 = L + 1.
    • L = (1 ± √5)/2. Every rn is positive, so L cannot be negative: L = (1 + √5)/2 = 1.618 (3 d.p.).
  4. r10 = 89/55; about 0.00015 above L
    • F10 = 55, F11 = 89, so r10 = 89/55 = 1.61818…
    • 1.61818… − 1.61803… = 0.000148…, which is 0.00015 (2 s.f.).

Task B: Periodic and patterned sequences

  1. −1, 1/2, 2: period 3
    • u2 = 1/(1 − 2) = −1
    • u3 = 1/(1 + 1) = 1/2
    • u4 = 1/(1 − 1/2) = 2 = u1, so the sequence repeats every 3 terms.
  2. 15
    • Each block of three terms adds to 2 − 1 + 1/2 = 3/2.
    • 30 terms make 10 blocks: 10 × 3/2 = 15
  3. 67
    • Fn is even exactly when 3 divides n: 33 of the first 100.
    • 100 − 33 = 67 are odd.

Task C: The golden ratio in a pentagon

  1. 108°
    • The interior angles add to (5 − 2) × 180° = 540°, so each is 540° ÷ 5 = 108°.
  2. 1.618
    • d2 = 12 + 12 − 2 × 1 × 1 × cos 108° = 2 − 2 cos 108° = 2.618…
    • d = 1.618 (4 s.f.): the golden ratio.
  3. d = 2 cos 36° = (1 + √5)/2 = φ
    • The diagonal is the base of an isosceles triangle with two sides 1 and apex angle 108°, so its base angles are 36°.
    • Dropping a perpendicular from the apex splits the base into two halves of length cos 36°, so d = 2 cos 36°.
    • 2 × (1 + √5)/4 = (1 + √5)/2 = φ

Extension

  1. A = 1/√5, B = −1/√5
    • φn+2 = φnφ2 = φn(φ + 1) = φn+1 + φn, and the same for ψ, so any combination works.
    • Aφ + Bψ = 1 and Aφ2 + Bψ2 = 1. Subtracting gives A + B = 0 (because φ2 − φ = ψ2 − ψ = 1).
    • So A(φ − ψ) = 1, and φ − ψ = √5: A = 1/√5 and B = −1/√5.
  2. F25 = 75 025
    • |ψn/√5| < 1/√5 < ½ for every n ≥ 1, so Fn is within ½ of φn/√5.
    • φ25/√5 = 75 024.999997…, so F25 = 75 025.

The same answers are in the answers PDF. Every answer was recomputed by computer before publishing.

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