Mathematics Advanced • Year 12 • Module 7 • Lesson 11
Recurrence Relations for Investments
Practise HSC-style writing on investment recurrences and an extended response on recurrence-vs-closed-form choices.
1. Short-answer questions
1.1 Sara invests $5,000 in an account paying 4.8% p.a. compounded monthly. She also deposits $200 at the end of each month.
(a) Write the recurrence relation An+1 in terms of An.
(b) Find A2 to the nearest cent. 3 marks Band 3
1.2 An investment satisfies the recurrence An+1 = 1.006An + 150 with A0 = $3,000.
(a) Use the closed form to find A3.
(b) State the difference (in cents) between the closed-form value and the value found by iteration, and explain in one sentence why the two methods agree to within rounding. 3 marks Band 3-4
1.3 A savings strategy starts with A0 = $0 and deposits a = $500 at the end of each year at 6% p.a. compounded annually.
(a) Write a recurrence relation for An+1.
(b) Find A3 by iteration.
(c) Use the closed form to find A10. 4 marks Band 4
2. Extended response
2.1 An investor opens an account with A0 = $10,000. Interest is 6% p.a. compounded monthly. The investor deposits $200 at the end of every month.
(a) Write the monthly recurrence relation for An+1, stating the periodic rate r used and explaining how it relates to the headline 6% p.a.
(b) Use iteration to compute A3 to the nearest cent.
(c) Use the closed-form formula to compute A120 (10 years) to the nearest cent.
(d) Identify one realistic situation where the bank must use the recurrence rather than the closed form to update the balance, and explain why in 2-3 sentences. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — correct recurrence An+1 = 1.005An + 200.
• 1 mark — explicitly identifies r = 0.06 ÷ 12 = 0.005 per month and states that the monthly rate is obtained from the annual rate by dividing by 12.
Part (b) — 2 marks
• 1 mark — at least one correct intermediate step shown (A1 or A2).
• 1 mark — correct final A3 to the nearest cent.
Part (c) — 2 marks
• 1 mark — correct substitution into An = A0(1+r)n + a × [(1+r)n − 1] ÷ r.
• 1 mark — correct numerical answer to the nearest cent.
Part (d) — 1 mark
• 1 mark — identifies a situation with variable cash flows (e.g. extra deposit, withdrawal, rate change, fees) and explains that the closed form assumes constant a and r, so it cannot accommodate the variation while the recurrence updates step by step.
Your response:
Stuck on (d)? Real banks face changing contributions (bonuses, missed deposits) or changing rates — both break the closed-form assumption of constants.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Sara's $5,000 + $200 account (3 marks)
Sample response. (a) r = 0.048 ÷ 12 = 0.004 per month, so An+1 = 1.004An + 200. (b) A1 = 1.004(5,000) + 200 = 5,020 + 200 = $5,220.00. A2 = 1.004(5,220) + 200 = 5,240.88 + 200 = $5,440.88.
Marking notes. 1 mark — correct recurrence with the right coefficient 1.004. 1 mark — correct A1. 1 mark — correct A2. A common error is writing 1.048An instead of 1.004An (failing to divide by 12) — this scores 0/3 on the calculation.
1.2 — Recurrence with An+1 = 1.006An + 150 (3 marks)
Sample response. (a) (1.006)³ = 1.018107. A3 = 3,000(1.018107) + 150 × (0.018107) ÷ 0.006 = 3,054.32 + 452.68 = $3,507.00 (closed form). By iteration: A1 = $3,168, A2 = $3,337.01, A3 = $3,507.03. (b) Difference = $0.03. The two methods are algebraically equivalent — the recurrence rounds at each step (3 small roundings), the closed form rounds only once, so a few cents may differ.
Marking notes. 1 mark — correct (1.006)³ value. 1 mark — correct closed-form numerical answer to nearest cent. 1 mark — correctly states the difference in cents and explains rounding. A response that simply restates "both should match" with no rounding insight scores 0 on the explanation part.
1.3 — A0 = 0, a = 500, r = 6% annually (4 marks)
Sample response. (a) An+1 = 1.06An + 500. (b) A1 = 1.06(0) + 500 = $500. A2 = 1.06(500) + 500 = $1,030. A3 = 1.06(1,030) + 500 = $1,591.80. (c) Closed form with A0 = 0: A10 = 500 × [(1.06)¹⁰ − 1] ÷ 0.06 = 500 × (1.790847 − 1) ÷ 0.06 = 500 × 13.18079 = $6,590.40.
Marking notes. 1 mark — correct recurrence. 1 mark — correct A3 by iteration. 1 mark — correct substitution into closed form. 1 mark — correct A10 value to the nearest cent. Common error: including a starting-balance term A0(1.06)¹⁰ when A0 = 0 — harmless here but a marker may deduct if the working obscures the answer.
2.1 — $10,000 + $200/month at 6% p.a. monthly (7 marks): sample Band-6 response
Sample Band-6 response.
(a) Recurrence. The monthly rate r = 0.06 ÷ 12 = 0.005, obtained by dividing the annual rate by the number of compounding periods per year. The recurrence is:
An+1 = 1.005 × An + 200, with A0 = 10,000.
[2 marks]
(b) Iteration to A3.
A1 = 1.005(10,000) + 200 = 10,050 + 200 = $10,250.00
A2 = 1.005(10,250) + 200 = 10,301.25 + 200 = $10,501.25
A3 = 1.005(10,501.25) + 200 = 10,553.76 + 200 = $10,753.76
[2 marks]
(c) Closed form to A120. (1.005)¹²⁰ = 1.819397. Then:
A120 = 10,000(1.819397) + 200 × (1.819397 − 1) ÷ 0.005
= 18,193.97 + 200 × 163.8794
= 18,193.97 + 32,775.88 = $50,969.85.
[2 marks]
(d) When the bank must use the recurrence. If the investor makes a one-off extra deposit of $1,000 in month 24, the regular contribution a is no longer constant — the closed form An = A0(1+r)n + a × [(1+r)n − 1] ÷ r assumes the same a in every period. The bank therefore uses the recurrence, applying the larger deposit at month 24 and the regular $200 in every other month, so the balance reflects the actual cash flows. [1 mark]
Total: 7/7.
Band descriptors for marker.
Band 3: Writes a recurrence that is approximately correct (right structure, possibly wrong r). Iterates to A3 with at least one correct step. Closed form attempted but with arithmetic errors. ≈ 3 marks.
Band 4: Correct recurrence, correct A3, correct A120 but no insight in (d) — just "the recurrence is easier" or similar. ≈ 4-5 marks.
Band 5: Full calculations correct. Part (d) identifies that the closed form assumes constants but does not name a specific real-world trigger. ≈ 6 marks.
Band 6: All calculations correct to the cent. Part (d) names a concrete trigger (one-off deposit, missed payment, rate change, fee, withdrawal) and explicitly links it to the constancy assumption in the closed-form formula. 7/7.