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.