Mathematics Advanced • Year 12 • Module 7 • Lesson 11

Recurrence Relations for Investments

Build fluency writing and iterating the recurrence A_n+1 = (1+r)A_n + a, and verifying with the closed form.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Complete the investment recurrence (with regular contribution a):

A_n+1 = ____________________

Q1.2 An account pays 6% p.a. compounded monthly. State the periodic rate r and the number of periods n for a 5-year term.

r = ____________ n = ____________

Q1.3 Write the closed-form formula for A_n when contributions a are made each period:

Stuck? Revisit lesson § Formula Reference and § Closed Form.

2. Worked example — Iterate, then verify

Follow every line. Each step has a short reason.

Problem. A₀ = $2,000, r = 0.6% per month (0.006), a = $150 deposited at the end of each month. Find A₄ by iteration and verify with the closed form.

Step 1 — Write the recurrence.

A_n+1 = 1.006 × A_n + 150

Reason: each period the previous balance earns interest, then the contribution is added.

Step 2 — Iterate to A₄.

A₁ = 1.006(2,000) + 150 = 2,012.00 + 150 = $2,162.00

A₂ = 1.006(2,162.00) + 150 = 2,174.97 + 150 = $2,324.97

A₃ = 1.006(2,324.97) + 150 = 2,338.92 + 150 = $2,488.92

A₄ = 1.006(2,488.92) + 150 = 2,503.85 + 150 = $2,653.85

Step 3 — Verify with the closed form.

A₄ = 2,000(1.006)⁴ + 150 × [(1.006)⁴ − 1] ÷ 0.006

= 2,000(1.024217) + 150 × (4.036125)

= 2,048.43 + 605.42 = $2,653.85

Conclusion. Both methods give A₄ = $2,653.85 — the recurrence and the closed form agree.

3. Faded example — fill in the missing steps

A₀ = $3,000, r = 0.5% per month, a = $100. Fill in each blank line. 4 marks

Step 1 — Recurrence:

A_n+1 = __________ × A_n + __________

Step 2 — Iterate.

A₁ = 1.005(3,000) + 100 = ______________ + 100 = $______________

A₂ = 1.005(______________) + 100 = $______________

A₃ = 1.005(______________) + 100 = $______________

Step 3 — Verify A₃ by closed form.

A₃ = 3,000(1.005)³ + 100 × [(1.005)³ − 1] ÷ 0.005 = ______________ + ______________ = $______________

Conclusion. After 3 months the balance is $______________. The recurrence and closed-form values agree to within rounding.

Stuck? Revisit lesson § Worked Example — Try It Now.

4. Graduated practice — write or iterate the recurrence

Show the substitution and the final value (to nearest cent unless stated). Use the same time units for r and n.

Foundation — single-step substitution (4 questions)

Q	Scenario	Working & answer
4.1 1	Write the recurrence for A₀ = $1,000, r = 0.4% per month, a = $50.
4.2 1	Given A₀ = $5,000 and A_n+1 = 1.004A_n + 200, find A₁.
4.3 1	Same as 4.2 — find A₂.
4.4 1	An account pays 7.2% p.a. compounded monthly. State r per month and n for 4 years.

Standard — typical HSC difficulty (6 questions)

Show at least one substitution line and one evaluation line.

4.5 A₀ = $4,000, r = 0.5% per month, a = $200. Find A₃ by iteration. 2 marks

4.6 An investment of $10,000 earns 6% p.a. compounded annually with no extra contributions. Use the closed form A_n = A₀(1+r)ⁿ to find A₅. 2 marks

4.7 A₀ = $0, r = 0.5% per month, a = $300. Find A₁₂ using the closed form. (No starting balance means only the contribution term contributes.) 2 marks

4.8 A₀ = $8,000, r = 0.5% per month, a = $250. Write the recurrence and find A₂ by iteration. 2 marks

4.9 A₀ = $5,000, r = 0.4% per month, a = $200. Use the closed form to find A₂₄ (2 years). 2 marks

4.10 An investment earns 6% p.a. compounded monthly. A₀ = $1,000, a = $100. State the recurrence and find A₆ by iteration. 2 marks

Extension — combine concepts (2 questions)

4.11 Two strategies grow for 20 years at r = 6% p.a. compounded annually. Strategy A: A₀ = $5,000, a = $0. Strategy B: A₀ = $0, a = $300/year. Find both A₂₀ values and state which wins and by how much. 3 marks

4.12 A₀ = $2,000, r = 0.005 per month, a = $100. Iterate to A₃, then verify by the closed form. State the rounding-induced difference (if any) in cents. 3 marks

Stuck on 4.11? The Strategy B answer uses the annuity term a × [(1+r)ⁿ − 1]/r alone.

5. Self-check the easy 3

Tick the first three once you have checked the method works.

For 4.1 my recurrence is A_n+1 = 1.004A_n + 50.

For 4.2 I got A₁ = 1.004(5,000) + 200 = $5,220.00.

For 4.4 I obtained r = 0.006 and n = 48 — both expressed in months.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Recurrence

A_n+1 = (1 + r)A_n + a.

Q1.2 — Monthly rate and periods

r = 0.06 ÷ 12 = 0.005 per month. n = 5 × 12 = 60 months.

Q1.3 — Closed form

A_n = A₀(1 + r)ⁿ + a × [(1 + r)ⁿ − 1] ÷ r.

Q3 — Faded example: A₀ = 3,000, r = 0.005, a = 100

Recurrence: A_n+1 = 1.005A_n + 100.
A₁ = 1.005(3,000) + 100 = 3,015 + 100 = $3,115.00.
A₂ = 1.005(3,115) + 100 = 3,130.58 + 100 = $3,230.58.
A₃ = 1.005(3,230.58) + 100 = 3,246.73 + 100 = $3,346.73.
Closed form: 3,000(1.005)³ + 100 × [(1.005)³ − 1] ÷ 0.005 = 3,045.23 + 301.51 = $3,346.74 (matches within $0.01 rounding).

Q4.1 — Recurrence for A₀ = 1,000, r = 0.004, a = 50

A_n+1 = 1.004A_n + 50.

Q4.2 — A₁ when A_n+1 = 1.004A_n + 200 and A₀ = 5,000

A₁ = 1.004(5,000) + 200 = 5,020 + 200 = $5,220.00.

Q4.3 — A₂

A₂ = 1.004(5,220) + 200 = 5,240.88 + 200 = $5,440.88.

Q4.4 — 7.2% monthly, 4 years

r = 0.072 ÷ 12 = 0.006 per month. n = 4 × 12 = 48 months.

Q4.5 — A₀ = 4,000, r = 0.005, a = 200, find A₃

A₁ = 1.005(4,000) + 200 = 4,020 + 200 = $4,220.00.
A₂ = 1.005(4,220) + 200 = 4,241.10 + 200 = $4,441.10.
A₃ = 1.005(4,441.10) + 200 = 4,463.31 + 200 = $4,663.31.

Q4.6 — $10,000 at 6% p.a., 5 years, no contribution

A₅ = 10,000(1.06)⁵ = 10,000 × 1.338226 = $13,382.26.

Q4.7 — A₀ = 0, r = 0.005, a = 300, n = 12

A₁₂ = 300 × [(1.005)¹² − 1] ÷ 0.005 = 300 × (1.061678 − 1) ÷ 0.005 = 300 × 12.33556 = $3,700.67.

Q4.8 — A₀ = 8,000, r = 0.005, a = 250, find A₂

Recurrence: A_n+1 = 1.005A_n + 250.
A₁ = 1.005(8,000) + 250 = 8,040 + 250 = $8,290.00.
A₂ = 1.005(8,290) + 250 = 8,331.45 + 250 = $8,581.45.

Q4.9 — A₀ = 5,000, r = 0.004, a = 200, n = 24

(1.004)²⁴ = 1.100530. A₂₄ = 5,000(1.100530) + 200 × (1.100530 − 1) ÷ 0.004 = 5,502.65 + 200 × 25.1325 = 5,502.65 + 5,026.51 = $10,529.16.

Q4.10 — 6% p.a. monthly, A₀ = 1,000, a = 100, find A₆

r = 0.005 per month. Recurrence: A_n+1 = 1.005A_n + 100.
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.
A₄ = 1.005(1,316.58) + 100 = $1,423.16.
A₅ = 1.005(1,423.16) + 100 = $1,530.28.
A₆ = 1.005(1,530.28) + 100 = $1,637.93.

Q4.11 — Strategy A vs Strategy B at 6% for 20 years

Strategy A: A₂₀ = 5,000(1.06)²⁰ = 5,000 × 3.207135 = $16,035.68.
Strategy B: A₂₀ = 300 × [(1.06)²⁰ − 1] ÷ 0.06 = 300 × 36.7856 = $11,035.68.
Strategy A wins by 16,035.68 − 11,035.68 = $5,000.00. A lump sum compounding for the full term beats $300/year that has not been deposited yet for most of the term.

Q4.12 — A₀ = 2,000, r = 0.005, a = 100, find A₃ two ways

Recurrence. A₁ = 1.005(2,000) + 100 = $2,110.00. A₂ = 1.005(2,110) + 100 = $2,220.55. A₃ = 1.005(2,220.55) + 100 = $2,331.65.
Closed form. A₃ = 2,000(1.005)³ + 100 × [(1.005)³ − 1] ÷ 0.005 = 2,030.15 + 301.51 = $2,331.66.
Difference = $0.01 due to intermediate rounding — methods agree.