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Themed maths · Mid-October

Ada Lovelace Day maths for IB Maths SL and HL

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 does the same with sequences: a 5-minute starter, a 35-minute main activity on algorithms and loops, an extension on Bernoulli numbers and full worked answers.

Level
IB Maths SL and HL (AA and AI)
Time
40 minutes, plus a 10-minute extension
Topics
Arithmetic and geometric sequences; Sigma notation and series; Recursive definitions and algorithms; Proof by induction (HL); Bernoulli numbers (extension)
Equipment
The starter is non-calculator. A GDC helps in task B.

Download student sheet (PDF)Answers (PDF)

Suggested timings

PartTimeWhat
Starter5 minQuick questions on the board
Main: task A10 minTrace the algorithm
Main: task B12 minLoops that make sequences
Main: task C13 minA recursive rule
Extension10 minFast finishers or homework

Starter (5 minutes)

No GDC. Four quick questions.

  1. An arithmetic sequence has first term 3 and common difference 5. Find the 20th term.
  2. Find the value of Σr=110 (2r + 1).
  3. Find the 6th term of the geometric sequence 2, 6, 18, …
  4. A loop starts with x = 1 and repeats the instruction ‘replace x by 2x + 1’ four times. What is x at the end?

Main activity (35 minutes)

Task A: Trace the algorithm (10 min)

Step 1: set n = 1 and S = 0. Step 2: add n2 to S. Step 3: if n = 10, output S and stop; otherwise increase n by 1 and go back to step 2.

  1. What number does the algorithm output?
  2. Write the output in sigma notation. Find the output when ‘n = 10’ in step 3 is changed to ‘n = 20’.
  3. With the stopping condition ‘n = 20’, how many times is step 2 carried out?

Task B: Loops that make sequences (12 min)

Program P sets u = 5, then repeats: ‘print u, then add 4 to u’. Program Q sets u = 1000, then repeats: ‘print u, then multiply u by 0.9’.

  1. Find the 15th number printed by program P.
  2. Find the sum of the first 15 numbers printed by program P.
  3. Which number printed by program Q is the first one below 100? Give its position and its value to 3 significant figures.
  4. If program Q ran for ever, what would the total of all the numbers it prints approach?

Task C: A recursive rule (13 min)

A sequence is defined by u1 = 1 and un+1 = 2un + 1.

  1. Find u2, u3, u4 and u5.
  2. Suggest a formula for un and use it to find u10.
  3. Find u1 + u2 + … + u10.

Extension (10 minutes)

Extension 1 is for everyone; extension 2 is HL (proof by induction).

  1. The Bernoulli numbers B0, B1, B2, … can be defined by B0 = 1 and, for m ≥ 1, Σk=0m m+1Ck Bk = 0. Use this rule, one value of m at a time, to find B1, B2, B3 and B4.
  2. (HL) Prove by induction that the sequence in task C has un = 2n − 1 for every positive integer n.

For teachers

Teacher notes and full worked answers
  • Ask students to write task A as a flowchart or in pseudocode first: tracing an algorithm by hand is the skill.
  • In task B (c), many students give 22: check whether they count the first number printed as number 1 or number 0.
  • Ada Lovelace is widely known for her notes on the Analytical Engine; one of them sets out a method for the Bernoulli numbers in extension 1. Keep the lesson about the maths.

Starter

  1. 98
    • u20 = 3 + 19 × 5 = 98
  2. 120
    • 2(1 + 2 + … + 10) + 10 = 2 × 55 + 10 = 120
  3. 486
    • u6 = 2 × 35 = 2 × 243 = 486
  4. 31
    • 1 → 3 → 7 → 15 → 31

Task A: Trace the algorithm

  1. 385
    • It adds 12 + 22 + … + 102.
    • Using Σr2 = n(n + 1)(2n + 1)/6: 10 × 11 × 21 ÷ 6 = 385
  2. Σr=120 r2 = 2870
    • 20 × 21 × 41 ÷ 6 = 2870
  3. 20 times
    • Step 2 runs once for each of n = 1, 2, …, 20: 20 times.

Task B: Loops that make sequences

  1. 61
    • The printed numbers are 5, 9, 13, …: arithmetic with d = 4.
    • u15 = 5 + 14 × 4 = 61
  2. 495
    • S15 = 15/2 × (5 + 61) = 495
  3. The 23rd number, 98.5
    • The kth number printed is 1000 × 0.9k−1.
    • 1000 × 0.9k−1 < 100 gives 0.9k−1 < 0.1, so k − 1 > ln 0.1 ÷ ln 0.9 = 21.85…
    • So k = 23, and 1000 × 0.922 = 98.47…, which is 98.5.
  4. 10 000
    • Geometric series with a = 1000 and r = 0.9.
    • S∞ = 1000 ÷ (1 − 0.9) = 10 000

Task C: A recursive rule

  1. 3, 7, 15, 31
    • 2 × 1 + 1 = 3, 2 × 3 + 1 = 7, 2 × 7 + 1 = 15, 2 × 15 + 1 = 31
  2. un = 2n − 1, so u10 = 1023
    • Each term is one less than a power of 2: 1 = 2 − 1, 3 = 4 − 1, 7 = 8 − 1, …
    • u10 = 210 − 1 = 1023
  3. 2036
    • Σ(2n − 1) = (2 + 4 + … + 210) − 10
    • The geometric part is 2(210 − 1) = 2046, so the total is 2046 − 10 = 2036.

Extension

  1. B1 = −1/2, B2 = 1/6, B3 = 0, B4 = −1/30
    • m = 1: B0 + 2B1 = 0, so B1 = −1/2.
    • m = 2: B0 + 3B1 + 3B2 = 0, so 1 − 3/2 + 3B2 = 0 and B2 = 1/6.
    • m = 3: 1 + 4(−1/2) + 6(1/6) + 4B3 = 0, so B3 = 0.
    • m = 4: 1 + 5(−1/2) + 10(1/6) + 10(0) + 5B4 = 0, so 5B4 = −1/6 and B4 = −1/30.
  2. Proof
    • Base case: u1 = 1 = 21 − 1.
    • Assume uk = 2k − 1 for some k. Then uk+1 = 2(2k − 1) + 1 = 2k+1 − 1.
    • It is true for n = 1, and if true for n = k it is true for n = k + 1, so by induction it is true for every positive integer n.

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

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