Mathematics Advanced • Year 12 • Module 7 • Lesson 11
Recurrence Relations for Investments
Apply investment recurrences to savings plans, strategy comparisons and design problems.
Problem 1 — The "$1,000 + $100" savings account (Think First scenario)
A savings account starts with $1,000. Each month it earns 0.5% interest and you deposit $100 at the end of the month.
Set up: What are we solving for?
(i) Write the recurrence An+1 in terms of An. 1 mark
(ii) Iterate to find A3 to the nearest cent. 2 marks
(iii) Verify A3 using the closed-form An = A0(1+r)n + a × [(1+r)n − 1] ÷ r. State the answers from each method and the rounding gap (cents). 3 marks
Stuck? Revisit lesson § Think First and § Closed Form.Problem 2 — Lump sum vs regular contributions
Two investors both invest at 6% p.a. compounded annually for 20 years.
Investor A: Lump sum A0 = $5,000, a = $0.
Investor B: A0 = $0, a = $300 deposited at the end of each year.
Set up: What are we solving for?
(i) Find A20 for Investor A using the closed form (no contributions). 2 marks
(ii) Find A20 for Investor B using the annuity term a × [(1+r)n − 1] ÷ r. 2 marks
(iii) State which investor wins and by how much. Then explain in one sentence why a lump sum beats equal-total regular contributions over the same horizon. 2 marks
Problem 3 — Designing a $50,000 savings plan
A student wants $50,000 in 10 years. They will earn 0.5% per month (6% p.a. monthly) on the account, starting with $10,000, and depositing a fixed $250 at the end of each month.
Set up: What are we solving for?
(i) Write the recurrence relation. 1 mark
(ii) Use the closed form to project A120 (10 years). Will the plan reach the $50,000 target? 3 marks
(iii) If the student stops contributing after 5 years (so a becomes 0 from month 60 onward), give the recurrence used for months 1–60 and the recurrence used for months 61–120. 2 marks
Stuck? Revisit lesson § Activity 3 — Design.Problem 4 — When does iteration beat the closed form?
A super fund posts the recurrence balance every day for a 35-year-old. The fund balance is currently $25,000, the daily growth rate is 0.025%, and the daily contribution is $40.
Set up: What are we solving for?
(i) Write the daily recurrence. 1 mark
(ii) Use the closed form to find the balance after 3 days. Show working. 2 marks
(iii) Explain in 2 sentences why the fund uses the recurrence (not the closed form) when daily contributions vary (e.g. an extra one-off bonus payment on day 90). 2 marks
Problem 5 — Catching a calculation error
A classmate claims that for A0 = $3,000, r = 0.006 per month, a = $150 per month, the balance after 4 months is A4 = $4,000.
Set up: What are we solving for?
(i) Use the recurrence to iterate to A4. Show every step. 3 marks
(ii) Verify A4 with the closed form. 2 marks
(iii) State by how much your answer differs from $4,000, and identify the most likely error your classmate made (e.g. omitting the (1+r) factor, using r as a percentage, miscounting contributions). 2 marks
Stuck? Revisit lesson § Misconceptions to Fix.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — The $1,000 + $100 account
Set up. Apply the recurrence An+1 = (1+r)An + a and verify with the closed form.
(i) An+1 = 1.005An + 100.
(ii) A1 = 1.005(1,000) + 100 = $1,105.00. A2 = 1.005(1,105) + 100 = $1,210.53. A3 = 1.005(1,210.53) + 100 = $1,316.58.
(iii) Closed form: A3 = 1,000(1.005)³ + 100 × [(1.005)³ − 1] ÷ 0.005 = 1,015.08 + 301.50 = $1,316.58. Recurrence and closed form agree to the cent.
Problem 2 — Lump sum vs contributions
Set up. Compute A20 for each strategy using the appropriate term of the closed form.
(i) Investor A: A20 = 5,000(1.06)²⁰ = 5,000 × 3.207135 = $16,035.68.
(ii) Investor B: A20 = 300 × [(1.06)²⁰ − 1] ÷ 0.06 = 300 × 36.7856 = $11,035.68.
(iii) Investor A wins by 16,035.68 − 11,035.68 = $5,000.00. Investor A's whole $5,000 compounds for the full 20 years, while Investor B's first $300 compounds only 19 years (and later contributions even fewer) — earlier dollars earn far more.
Problem 3 — $50,000 savings plan
Set up. Project a monthly recurrence for 10 years and judge against the $50,000 target.
(i) An+1 = 1.005An + 250.
(ii) (1.005)¹²⁰ = 1.819397. A120 = 10,000(1.819397) + 250 × (1.819397 − 1) ÷ 0.005 = 18,193.97 + 40,969.85 = $59,163.82. The plan exceeds the $50,000 target by $9,163.82.
(iii) Months 1–60: An+1 = 1.005An + 250. Months 61–120: An+1 = 1.005An + 0 (i.e. An+1 = 1.005An).
Problem 4 — Super fund daily recurrence
Set up. Iterate a daily recurrence then evaluate suitability of the closed form when contributions vary.
(i) An+1 = 1.00025An + 40.
(ii) A3 = 25,000(1.00025)³ + 40 × [(1.00025)³ − 1] ÷ 0.00025. (1.00025)³ = 1.0007502. A3 = 25,000(1.0007502) + 40 × 3.000750 = 25,018.75 + 120.03 = $25,138.78.
(iii) The closed form assumes a constant contribution a each period. As soon as one day's contribution differs (e.g. a bonus), the formula no longer applies. The recurrence updates from yesterday's balance using whatever contribution actually occurred today, so it handles variable cash flows naturally.
Problem 5 — Catching the error
Set up. Iterate the recurrence, then cross-check with the closed form and identify the likely mistake.
(i) A1 = 1.006(3,000) + 150 = 3,018 + 150 = $3,168.00. A2 = 1.006(3,168) + 150 = 3,187.01 + 150 = $3,337.01. A3 = 1.006(3,337.01) + 150 = 3,357.03 + 150 = $3,507.03. A4 = 1.006(3,507.03) + 150 = 3,528.07 + 150 = $3,678.07.
(ii) Closed form: (1.006)⁴ = 1.024217. A4 = 3,000(1.024217) + 150 × (0.024217) ÷ 0.006 = 3,072.65 + 605.42 = $3,678.07 — agrees with the recurrence.
(iii) The correct answer is $3,678.07, $321.93 short of $4,000. The most likely error is treating each month's deposit as itself accumulating instant interest (e.g. multiplying the contribution sum 4 × 150 = $600 by (1.006)⁴ separately) or doubling the contribution somewhere — both would inflate the answer by roughly $300, matching the gap.