Compound Interest Investments

Straight from the formula book. The QCAA General Mathematics formula book gives compound interest as A = P(1 + i)ⁿ, the recurrence relation as A_n+1 = rA_n where r = 1 + i, and the effective annual rate as i_effective = (1 + i)^k − 1. Those exact letters are used everywhere on this page so nothing surprises you in the exam.

What each letter means: A = total (final) amount · P = principal (starting money) · i = interest rate per compounding period, written as a decimal · n = number of compounding periods · r = the multiplier, 1 + i · k = compounding periods per year. Note the formula book uses i for the rate and r for the multiplier.

If you only do one thing Grow money with A = P(1 + i)ⁿ, then turn any account into one comparable number with i_effective = (1 + i)^k − 1. Higher effective rate wins. That is the whole topic.

The 10-minute version

Low energy? Do just this and you have still won the day. Zero is the only fail.

A = P (1 + i)ⁿ · i_effective = (1 + i)^k − 1

One example. $2,000 at 6% p.a. compounded quarterly for 2 years:

i (per period):6% ÷ 4 = 1.5% = 0.015

n (periods):4 × 2 = 8

A:2,000 × (1.015)⁸ ≈ $2,253

Then go straight to Practice Q1 and Q5. Screenshot the win for Nat and stop there.

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Watch it explained

Four short ways in. Pick whichever suits your energy. The 🎬 cards are waiting on Nat's videos, and each has a ready-to-read film script tucked underneath so the formula matches your formula book exactly.

The formula, and compounding frequency

Compound interest (formula book) A = P(1 + i)ⁿ
i = interest rate per compounding period (decimal) · n = number of compounding periods

In Unit 3 interest was usually calculated annually, so i was just the yearly rate and n was the number of years. In Unit 4 banks compound monthly, quarterly, even daily. The formula does not change, but i becomes the rate per period and n becomes the total number of periods.

Annual

Once per year

k = 1

Semi-annual

Twice per year

k = 2

Quarterly

4× per year

k = 4

Monthly

12× per year

k = 12

Convert before you substitute: i = annual rate ÷ k · n = k × number of years.
Example: 6% p.a. compounded monthly for 4 years gives i = 0.06 ÷ 12 = 0.005 and n = 12 × 4 = 48.

The recurrence relation form

The formula book also models compound interest as a recurrence relation, and the exam often asks for it by name (for example, "write a recurrence relation for the amount, where n is the number of months"). It just says "each period, multiply by the growth multiplier r".

Recurrence relation (formula book) A_n+1 = rA_n, where r = 1 + i
A₀ = principal · r = 1 + (rate per period) · each step is one compounding period

You must state the starting value A₀ as part of the answer. The closed formula A = P(1 + i)ⁿ is just this same rule applied n times in one line.

Worked examples

Worked Example A · annual compounding

$5,000 is invested at 6% p.a. compounded annually for 3 years. Find the value of the investment.

i (per period):6% p.a. ÷ 1 = 0.06

n (periods):1 × 3 = 3

Substitute:A = 5,000(1 + 0.06)³ = 5,000(1.06)³

A:5,000 × 1.191016 = $5,955.08

Worked Example B · exam style, monthly compounding

When a child is born, a parent deposits $3,000 into an account earning 4.2% p.a. compounding monthly. With no further transactions and no rate change, find the interest earned by the child's 18th birthday. (SEE 2024 Paper 1, Q18 style.)

i (per period):0.042 ÷ 12 = 0.0035

n (periods):12 × 18 = 216

Substitute:A = 3,000(1.0035)²¹⁶

A:3,000 × 2.126931 = $6,380.79

Interest:6,380.79 − 3,000 = $3,380.79

Read the question: it asks for interest earned, not the final balance. Find A first, then subtract the principal. Dropping this last line is the most common lost mark.

Worked Example C · same money, different frequency

$2,000 invested at 6% p.a. for 2 years. Compare the value under different compounding frequencies.

Compounding	k	i = 0.06 ÷ k	n = k × 2	A = 2000(1 + i)ⁿ
Annually	1	0.06	2	$2,247.20
Quarterly	4	0.015	8	$2,252.99
Monthly	12	0.005	24	$2,254.32

More frequent compounding gives a slightly higher A. The gap grows with higher rates and longer time.

Effective annual rate

The effective annual rate turns any compounding frequency into one equivalent annual rate, so you can compare accounts directly. The exam asks for it by this exact name, usually "as a percentage".

Effective annual rate (formula book) i_effective = (1 + i)^k − 1
i = interest rate per compounding period · k = compounding periods per year · answer is a decimal (× 100 for %)

It answers: "if this account compounded annually instead, what single rate would give the same outcome?" A higher effective rate is the better investment.

Worked Example D · comparing two accounts

Account A: 6% p.a. compounded quarterly (i = 0.015, k = 4). Account B: 6.1% p.a. compounded annually (i = 0.061, k = 1). Which has the higher effective annual rate?

Account A:i_eff = (1.015)⁴ − 1 = 1.06136 − 1 = 6.14%

Account B:i_eff = (1.061)¹ − 1 = 6.10%

Winner:Account A (6.14% > 6.10%), despite the lower advertised rate.

Key insight: a lower advertised (nominal) rate with more frequent compounding can beat a higher nominal rate. Always compare the effective rate, never the nominal rate.

Finding n, how long to reach a target?

If you know the target A and want to find when it is reached, rearrange the formula book rule using logarithms, or step up the table on your Casio with the Ans-key trick (the same one used for loans and savings tables).

Finding n using logarithms A = P(1 + i)ⁿ
→ n = log(A ÷ P) ÷ log(1 + i)
i is the rate per period · answer is a number of periods

Worked Example E · finding the time

How many years for $1,000 to grow to $2,197 at 30% p.a. compounded annually?

A ÷ P:2,197 ÷ 1,000 = 2.197

Log method:n = log(2.197) ÷ log(1.30)

Check:1.30¹ = 1.3 · 1.30² = 1.69 · 1.30³ = 2.197 ✓ → n = 3 years

When the numbers work out cleanly like this, checking by trial and error is often faster in an exam.

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How a marker wants it laid out

A future-value question written the QCAA way, using the formula book letters. Each line earns its own tick, so you bank marks even if the final figure slips.

Q. $2,000 is invested at 6% p.a. compounded quarterly for 2 years. Find its value.

i = 0.06 ÷ 4 = 0.015, n = 4 × 2 = 8 ✓ determines i and n

A = P(1 + i)ⁿ = 2000(1.015)⁸ ✓ substitutes into the rule

A = 2252.985… = $2252.99 ✓ computes the value

The investment is worth $2252.99 after 2 years. ✓ answer in context

If a question adds a claim, finish with a reasonableness line, e.g. "2252.99 > 2200, so it covers the $2200 cost." ✓ reasonableness

Practice Questions

1. Find the value of $5,000 invested at 6% p.a. compounded annually for 3 years. ▶

i = 0.06, n = 3 → A = 5,000(1.06)³ = 5,000 × 1.191016 = $5,955.08

2. $3,000 earns 4.2% p.a. compounding monthly. Write a recurrence relation for the amount A_n after n months, then use it to find the balance after 2 months. ▶

i = 0.042 ÷ 12 = 0.0035, so r = 1 + i = 1.0035.
Recurrence: A_n+1 = 1.0035 A_n, A₀ = 3000.
A₁ = 1.0035 × 3000 = $3,010.50
A₂ = 1.0035 × 3010.50 = $3,021.04

3. $2,000 is invested at 20% p.a. compounded annually for 2 years. Find the value and the interest earned. ▶

A = 2,000(1.20)² = 2,000 × 1.44 = $2,880
Interest = 2,880 − 2,000 = $880

4. $4,000 is invested at 10% p.a. compounded annually for 3 years. What is the total interest earned? ▶

i = 0.10, n = 3 → A = 4,000(1.10)³ = 4,000 × 1.331 = $5,324
Interest = 5,324 − 4,000 = $1,324

5. Two investments both start with $10,000: Plan X at 8% p.a. compounded annually, Plan Y at 7.8% p.a. compounded monthly. Which has the higher effective annual rate? ▶

Plan X: i = 0.08, k = 1 → i_eff = (1.08)¹ − 1 = 8.00%

Plan Y: i = 0.078 ÷ 12 = 0.0065, k = 12 → i_eff = (1.0065)¹² − 1 ≈ 8.09%

Plan Y wins. Despite the lower advertised rate, monthly compounding pushes its effective rate above Plan X.

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📈 Compound Interest Investments