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How does money grow when interest compounds, and how do regular savings build an investment?

Apply the compound interest formula and model annuities with regular contributions to find future values of investments.

How to use the compound interest formula with different compounding periods, find the future value of an investment, and model an annuity that grows through regular contributions.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The compound interest formula
Why compounding frequency matters
Annuities: investing with regular contributions
Comparing simple and compound interest
Solving for time or rate
Reading the result in context

What this dot point is asking

You must apply the compound interest formula with correct period rates, find future values, and model an annuity that grows through regular deposits.

The compound interest formula

Compound interest is calculated on the growing balance, so earlier interest itself earns interest.

A = P(1 + i)^n

Here $P$ is the principal (starting amount), $A$ is the final amount, $i$ is the interest rate per compounding period (as a decimal) and $n$ is the number of compounding periods.

The key step is converting an annual rate to the period rate and the years to the number of periods.

Worked example

$\$5000$ is invested at 6% per annum compounded monthly for 4 years. Find the final amount and the total interest earned.

Step 1: Convert the annual rate to a rate per compounding period

The formula $A = P(1 + i)^n$ requires $i$ and $n$ to use the same period. Because interest compounds monthly, we divide the annual rate by 12 to get the monthly rate:

i = \frac{0.06}{12} = 0.005.

This is the rate applied to the balance at the end of each month.

Step 2: Find the total number of compounding periods

Over 4 years with monthly compounding there are $4 \times 12 = 48$ compounding periods.

Step 3: Apply the compound interest formula

Now substitute $P = 5000$ , $i = 0.005$ and $n = 48$ into $A = P(1 + i)^n$ :

A = 5000(1.005)^{48} \approx 5000 \times 1.27049 \approx 6352.45.

Step 4: State the final amount and interest earned

The interest earned is the growth above the original principal:

\text{interest} = A - P \approx 6352.45 - 5000 = 1352.45.

Final answer: The investment grows to approximately $\$6352.45$ , earning about $\$1352.45$ in interest over 4 years.

Why compounding frequency matters

For the same annual rate, more frequent compounding gives slightly more growth, because interest is added to the balance sooner and starts earning interest itself. Daily beats monthly beats annual, though the differences are modest at typical rates.

Annuities: investing with regular contributions

An annuity is an investment that receives equal payments at regular intervals, with compound interest applied each period. The balance grows from two sources: interest on what is already there, plus each new contribution.

You can model an annuity step by step with a recurrence. If $V_n$ is the balance after $n$ periods, $i$ is the period rate and $d$ is the regular deposit made each period:

V_{n+1} = V_n(1 + i) + d

Worked example

A saver deposits $\$200$ at the end of each month into an account paying 6% per annum compounded monthly, starting from a zero balance. Find the balance after 3 months by applying the recurrence relation step by step.

Step 1: Identify the recurrence parameters

The recurrence $V_{n+1} = V_n(1 + i) + d$ models an annuity. We identify each quantity: the monthly interest rate is $i = \dfrac{0.06}{12} = 0.005$ , the regular deposit each month is $d = 200$ , and the starting balance is $V_0 = 0$ .

Step 2: Apply the recurrence for month 1

At the end of month 1, the existing balance earns interest and then the deposit is added. Because $V_0 = 0$ , no interest accrues this month and the balance is simply the first deposit:

V_1 = 0 \times (1.005) + 200 = 200.

Step 3: Apply the recurrence for month 2

At the end of month 2, the $\$200$ balance from month 1 earns one month's interest before the second deposit is added:

V_2 = 200(1.005) + 200 = 201 + 200 = 401.

Step 4: Apply the recurrence for month 3

At the end of month 3, the $\$401$ balance earns interest, then the third deposit arrives:

V_3 = 401(1.005) + 200 = 403.005 + 200 = 603.01.

The balance is slightly above $\$600$ because the earlier deposits have each earned a small amount of compound interest on top of the contributions.

Final answer: The balance after 3 months is approximately $\$603.01$ .

Comparing simple and compound interest

It helps to contrast compound interest with simple interest, where interest is computed only on the original principal: $A = P(1 + rt)$ . Simple interest grows in a straight line, while compound interest grows as a curve that steepens over time, because each period's interest is added to the balance and then earns interest itself. Over short periods the two are close, but over many years compound interest pulls well ahead. SACE problems sometimes ask you to compare the two over a given term, and the gap between them is precisely the "interest on the interest" that compounding generates.

Solving for time or rate

Beyond finding a final amount, you may be asked how long an investment takes to reach a target, or what rate is needed. For a lump sum, set $A = P(1 + i)^n$ equal to the target and solve - for the number of periods this needs logarithms, while a calculator's financial solver handles it directly. For an annuity, the financial solver with the present value, payment, rate and target future value finds the missing period count or rate in one step. Recognising which quantity is unknown and entering the rest into the solver is the practical skill the topic builds toward.

Reading the result in context

For investments, the future value answers "how much will I have?" and the interest earned is $A - P$ (or final balance minus total deposits for an annuity). Examiners often ask you to compare two options or to find how long an investment takes to reach a target.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20212 marksCalculator-assumed. Eilidh contributes approximately 1960 dollars per quarter to a superannuation fund earning 5.85% per annum, compounded quarterly. Calculate the expected balance after 43 years.

Show worked answer →

This is the future value of an annuity. Quarterly rate $i = \dfrac{0.0585}{4} = 0.014625$ ; periods $n = 43 \times 4 = 172$ ; payment $1960$ per quarter, present value $0$ .

FV = 1960 \times \frac{(1.014625)^{172} - 1}{0.014625} \approx 1\,080\,000.

So about 1.08 million dollars. Marks: one for the rate per period and number of periods, one for the future value of roughly 1.08 million.

SACE 20222 marksCalculator-assumed. Arjun invests 20000 dollars at 3.1% per annum compounded weekly and also deposits 50 dollars each week. Show the balance will be approximately 30100 dollars after 3 years.

Show worked answer →

Combine a compounding lump sum with a regular-deposit annuity. Weekly rate $i = \dfrac{0.031}{52}$ ; periods $n = 3 \times 52 = 156$ .

Lump sum: $20000(1 + i)^{156} \approx 21\,950$ . Deposits: $50 \times \dfrac{(1 + i)^{156} - 1}{i} \approx 8160$ .

Total $\approx 21\,950 + 8160 = 30\,110 \approx 30\,100$ dollars, as required. Marks: one for both components with the correct weekly rate, one for combining to about 30100 dollars.

SACE 20232 marksCalculator-assumed. Lewis saves 500 dollars per fortnight at 3.2% per annum compounded fortnightly. Show it takes approximately 62 fortnights to reach 32000 dollars.

Show worked answer →

Future value of an annuity, solving for $n$ . Fortnightly rate $i = \dfrac{0.032}{26} \approx 0.0012308$ ; set $FV = 32000$ , payment $500$ .

32000 = 500 \times \frac{(1 + i)^n - 1}{i}.

Solving with a financial solver gives $n \approx 61.8$ , so about 62 fortnights. Check: $62 \times 500 = 31\,000$ deposited plus about 1000 interest gives roughly 32000. Marks: one for the annuity set-up, one for solving to $n \approx 62$ .