A Level Math Revision Free diagnostic
Themed maths · Mid-October

Ada Lovelace Day maths for A Level

Ada Lovelace is widely known for her notes on the Analytical Engine, which set out step-by-step methods for a machine to follow. This lesson is about such methods: a 5-minute starter, a 35-minute main activity on algorithms, iteration and recurrence relations, an extension on Bernoulli numbers and full worked answers.

Level
A Level Maths (Year 12 and Year 13)
Time
40 minutes, plus a 10-minute extension
Topics
Sequences, series and sigma notation; Recurrence relations; Numerical methods: iteration; Geometric series in context; Bernoulli numbers (extension)
Equipment
The starter is non-calculator. A calculator is needed for tasks B and C.

Download student sheet (PDF)Answers (PDF)

Suggested timings

PartTimeWhat
Starter5 minQuick questions on the board
Main: task A10 minTrace the algorithm
Main: task B13 minAn algorithm that solves an equation
Main: task C12 minA savings algorithm
Extension10 minFast finishers or homework

Starter (5 minutes)

No calculator.

  1. Find Σr=120 r.
  2. A sequence has u1 = 2 and un+1 = 3un − 2. Find u3.
  3. Find the sum to infinity of the geometric series with first term 4 and common ratio ½.
  4. Which term of the sequence 1, 4, 7, 10, … is 100?

Main activity (35 minutes)

Task A: Trace the algorithm (10 min)

Step 1: set n = 1 and T = 0. Step 2: add n3 to T. Step 3: if n = 6, output T and stop; otherwise add 1 to n and go back to step 2.

  1. What does the algorithm output?
  2. Show that the output equals (1 + 2 + … + 6)2.
  3. Using Σr=1n r3 = ¼n2(n + 1)2, find the output when ‘n = 6’ is changed to ‘n = 30’.

Task B: An algorithm that solves an equation (13 min)

The equation x3 − 3x − 5 = 0 has one real root, α.

  1. Let f(x) = x3 − 3x − 5. Work out f(2) and f(3), and explain why α lies between 2 and 3.
  2. Use the iteration xn+1 = ∛(3xn + 5) with x1 = 2 to find x2, x3 and x4, each to 4 decimal places.
  3. Find α to 3 decimal places, and prove your answer is correct to that accuracy.

Task C: A savings algorithm (12 min)

A savings account starts with £500. At the end of each year the balance is multiplied by 1.03 and then £200 is added. Let Bn be the balance after n years, so B0 = 500 and Bn+1 = 1.03Bn + 200.

  1. Find B1 and B2.
  2. Use a geometric series to show that Bn = 500 × 1.03n + 200(1.03n − 1) ÷ 0.03, and find B10 to the nearest penny.
  3. After how many whole years is the balance first more than £5000?

Extension (10 minutes)

Sums of powers and the Bernoulli numbers.

  1. The Bernoulli numbers can be defined by B0 = 1 and, for m ≥ 1, Σk=0m m+1Ck Bk = 0. Find B1, B2, B3 and B4.
  2. The Bernoulli numbers give formulas for sums of powers, for example 14 + 24 + … + n4 = n5/5 + n4/2 + n3/3 − n/30. Check this formula when n = 3, then use it to find 14 + 24 + … + 104.

For teachers

Teacher notes and full worked answers
  • Task B is a good place to show a cobweb diagram on graphing software: the iteration converges because the gradient of ∛(3x + 5) near the root is less than 1.
  • In task C, students who use B1 as the starting balance get every later answer one year out: check the indexing first.
  • Ada Lovelace is widely known for her notes on the Analytical Engine; one of them sets out a method for the Bernoulli numbers in the extension. Keep the lesson about the maths.

Starter

  1. 210
    • 20 × 21 ÷ 2 = 210
  2. 10
    • u2 = 6 − 2 = 4, u3 = 12 − 2 = 10
  3. 8
    • 4 ÷ (1 − ½) = 8
  4. The 34th
    • un = 3n − 2 = 100 gives n = 34.

Task A: Trace the algorithm

  1. 441
    • 1 + 8 + 27 + 64 + 125 + 216 = 441
  2. 212 = 441
    • 1 + 2 + … + 6 = 21 and 212 = 441, the same as the output.
  3. 216 225
    • ¼ × 302 × 312 = ¼ × 900 × 961 = 216 225

Task B: An algorithm that solves an equation

  1. f(2) = −3, f(3) = 13: a change of sign
    • f(2) = 8 − 6 − 5 = −3 and f(3) = 27 − 9 − 5 = 13.
    • f is continuous and changes sign, so there is a root between 2 and 3.
  2. 2.2240, 2.2684, 2.2770
    • x2 = ∛11 = 2.22398…
    • x3 = ∛(3 × 2.22398… + 5) = 2.26837…
    • x4 = ∛(3 × 2.26837… + 5) = 2.27696…
  3. α = 2.279
    • Carrying on the iteration gives 2.2790…
    • f(2.2785) = −0.0065… < 0 and f(2.2795) = 0.0061… > 0.
    • The sign changes, so 2.2785 < α < 2.2795 and α = 2.279 to 3 d.p.

Task C: A savings algorithm

  1. £715 and £936.45
    • 1.03 × 500 + 200 = 715
    • 1.03 × 715 + 200 = 936.45
  2. £2964.73
    • The first £500 has grown to 500 × 1.03n. The £200 payments have grown to 200(1 + 1.03 + … + 1.03n−1) = 200(1.03n − 1) ÷ 0.03.
    • B10 = 500 × 1.0310 + 200(1.0310 − 1) ÷ 0.03 = 2964.734…
  3. 17 years
    • Solve 500 × 1.03n + 200(1.03n − 1) ÷ 0.03 > 5000, which gives 1.03n > 11666.67 ÷ 7166.67 = 1.6279…
    • n > ln 1.6279 ÷ ln 1.03 = 16.49…, so 17 years (B16 = 4882.27, B17 = 5178.74).

Extension

  1. −1/2, 1/6, 0, −1/30
    • m = 1: 1 + 2B1 = 0, so B1 = −1/2.
    • m = 2: 1 + 3(−1/2) + 3B2 = 0, so B2 = 1/6.
    • m = 3: 1 − 2 + 1 + 4B3 = 0, so B3 = 0.
    • m = 4: 1 − 5/2 + 10/6 + 5B4 = 0, so B4 = −1/30.
  2. 98 when n = 3; 25 333
    • n = 3: 243/5 + 81/2 + 9 − 1/10 = 48.6 + 40.5 + 9 − 0.1 = 98, and 1 + 16 + 81 = 98.
    • n = 10: 20 000 + 5000 + 1000/3 − 1/3 = 25 333

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

Practise the topics

Halloween maths · Bonfire Night maths · Fibonacci Day maths · All themed maths

More for lessons: Weekly starters · Skill Builders · Worksheet builder · Competition maths