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.