Compound interest with recursion in WACE Mathematics Applications Unit 3

Syllabus dot point

“Model compound interest with a recurrence relation, convert nominal annual rates to the period rate, and find balances and effective rates.”

WACEMathematics ApplicationsUnit 38 min readUpdated 2 Jun 2026

Quick answer

Compound interest grows a balance by the same percentage each period, so it is geometric. Model it with the recurrence $A_{n+1} = R\,A_n$ where $R = 1 + i$ and $i$ is the rate per compounding period. Convert a nominal annual rate to the period rate by dividing by the number of periods per year, then count periods, not years. Compare investments fairly with the effective annual rate $(1+i)^k - 1$ .

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What this dot point is asking
The compound interest recurrence
Converting nominal rates
The effective annual rate
Using the recurrence on a calculator
Exam-style questions

What this dot point is asking

You must set up the recurrence, convert nominal rates to the per-period rate, find any balance, and calculate and compare effective annual rates.

The compound interest recurrence

Each period the balance is multiplied by a fixed growth factor, so it is a geometric model.

Compound interest recurrence

$A_{n+1} = R\,A_n$ , where $R = 1 + i$ and $i$ is the interest rate per compounding period (as a decimal). The explicit form is $A_n = A_0 R^n$ , where $A_0$ is the principal and $n$ is the number of compounding periods.

The key word is period. If interest compounds monthly, $n$ counts months and $i$ is the monthly rate; if it compounds quarterly, $n$ counts quarters. Mixing years with a monthly rate is the most common error in the topic.

Converting nominal rates

A nominal annual rate is the headline yearly rate. The rate actually applied each period is the nominal rate divided by the number of compounding periods per year.

Compounding monthly: $i = \dfrac{\text{nominal}}{12}$ , and $n$ counts months.
Compounding quarterly: $i = \dfrac{\text{nominal}}{4}$ , and $n$ counts quarters.
Compounding daily: $i = \dfrac{\text{nominal}}{365}$ , and $n$ counts days.

For a nominal $7.2\%$ per annum compounding monthly, $i = \dfrac{0.072}{12} = 0.006$ , so $R = 1.006$ .

Periods, not years

Always raise the growth factor to the number of compounding periods. A $5$ year investment compounding quarterly uses $n = 5 \times 4 = 20$ , not $n = 5$ . Count the periods first, then substitute.

Worked example

Finding a balance and the interest earned

$\$12\,000$ is invested at $6\%$ per annum compounding quarterly for $3$ years. Find the balance and the total interest.

Step 1: Convert the nominal rate to the period rate

The nominal annual rate must be divided by the number of compounding periods per year to get the rate that actually applies each period. Quarterly rate $i = \dfrac{0.06}{4} = 0.015$ , so the growth factor $R = 1.015$ .

Step 2: Count the total number of compounding periods

Because interest is applied quarterly, time must be counted in quarters rather than years. Three years compounding quarterly gives $n = 3 \times 4 = 12$ periods.

Step 3: Apply the explicit rule to find the balance

Raising the growth factor to the number of periods gives the balance directly from the principal. $A_{12} = 12000 \times 1.015^{12} = 12000 \times 1.19562 = 14347.45$ , so the balance is $\$14\,347.45$ .

Step 4: Find the total interest

The interest earned is the final balance minus the original principal. Interest $= 14347.45 - 12000 = 2347.45$ , so $\$2347.45$ in total. Notice the interest is more than $6\% \times 3 = 18\%$ of the principal because each quarter earns interest on the interest already added.

Final answer: Balance after $3$ years is $\$14\,347.45$ ; total interest earned is $\$2347.45$ .

The effective annual rate

Two accounts with the same nominal rate are not equally good if they compound at different frequencies. The effective annual rate restates a nominal rate as the equivalent rate compounding once a year, so accounts can be compared fairly.

Effective annual rate

$i_{\text{eff}} = (1 + i)^k - 1$ , where $i$ is the per-period rate and $k$ is the number of compounding periods per year.

For a nominal $6\%$ compounding monthly, $i = 0.005$ and $i_{\text{eff}} = 1.005^{12} - 1 = 0.06168 = 6.17\%$ . The more frequent the compounding, the higher the effective rate, because interest is earned on interest sooner.

Using the recurrence on a calculator

To generate balances term by term, store the principal, then repeatedly multiply by $R$ . This is how the SCSA calculator-assumed section expects you to verify a balance or find when an investment first passes a target. To find the first $n$ for which $A_n$ exceeds, say, $\$20\,000$ , increase $n$ until the balance crosses the threshold, then report the first whole period above it.

Common errors

Using the nominal rate per period: Divide the nominal annual rate by the number of periods per year before using it; a monthly model never uses the annual percentage directly.
Counting years instead of periods: Raise $R$ to the number of compounding periods, so a quarterly investment over $4$ years uses $n = 16$ .
Confusing nominal and effective rates: The nominal rate is the headline; the effective rate accounts for compounding frequency. Compare accounts on effective rates.
Rounding mid-calculation: Keep full precision in $R^n$ and round only the final dollar amount, or accumulated rounding shifts the answer by cents or more.

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2021-style6 marks

\$8000

is invested at

4.8\%

per annum compounding monthly. (a) Write a recurrence relation for the balance

A_n

after

n

months. (b) Find the balance after

2

years. (c) Find the effective annual interest rate, correct to two decimal places.

Show worked answer →

Compounding monthly means the monthly rate is the nominal annual rate divided by $12$ .

(a) Monthly rate $= \dfrac{4.8\%}{12} = 0.4\% = 0.004$ . Recurrence: $A_{n+1} = 1.004\,A_n$ , with $A_0 = 8000$ . (2 marks)

(b) Two years is $24$ months, so $A_{24} = 8000 \times 1.004^{24} = 8000 \times 1.10057 = 8804.55$ , that is $\$8804.55$ . (2 marks)

Markers reward the per-period rate, the correct number of periods, and an effective rate from one year of compounding.

WACE 2023-style4 marks

Two banks offer the same nominal rate of

6\%

per annum. Bank A compounds annually and Bank B compounds quarterly. Show that Bank B gives the higher effective annual rate and state the difference correct to two decimal places.

Show worked answer →

Compare the growth factor over one year for each bank.

Bank A effective rate $= 6.00\%$ because it compounds once a year. Bank B quarterly rate $= \dfrac{6\%}{4} = 1.5\% = 0.015$ , so its effective rate $= 1.015^4 - 1 = 1.06136 - 1 = 0.06136 = 6.14\%$ . (3 marks)

Bank B is higher by $6.14\% - 6.00\% = 0.14\%$ , because more frequent compounding earns interest on interest sooner. (1 mark)

Markers reward the quarterly rate, the one-year growth factor raised to the fourth power, and the correct difference.