Each period you earn interest on your money and on the interest you already earned. So it snowballs: every period's interest is a little bigger than the last.
A = total amount at the end · P = principal (starting money) · i = interest rate per period (as a decimal) · n = number of periods.
If it compounds more than once a year, first find i, the rate per period: annual rate ÷ periods per year. (6% compounding monthly = 0.5% = 0.005 per month.)
The formula book also writes this as a recurrence relation: An+1 = r An, where r = 1 + i and A0 is the principal. Same rule, one step at a time.
$5,000 invested at 6% p.a., compounding annually, for 3 years.
| End of year | Calculation | Balance |
|---|---|---|
| Start (A0) | opening | $5,000.00 |
| 1 | 5000 × 1.06 | $5,300.00 |
| 2 | 5300 × 1.06 | $5,618.00 |
| 3 | 5618 × 1.06 | $5,955.08 |
• Use i, the rate per period, not always the yearly one (divide by the periods per year).
• Round only at the very end, not at each step.
• If a question asks for interest, subtract the principal: interest = A − P.
• Compound is not simple interest: the interest grows each period.
Compound interest is just "times the multiplier r = 1 + i", once for each period. That is the recurrence An+1 = r An.
$5,000 at 6%. Simple interest adds the same $300 every year (straight line). Compound earns interest on the interest, so the line curves up and away.
Don't read yet, just have a go in your head:
i = 0.05, n = 3, r = 1.05 → A = 2000(1.05)3: 2000 × 1.05 = $2,100 → × 1.05 = $2,205 → × 1.05 = $2,315.25
r = 1.05 → 2000 × 1.05 = $2,100 → × 1.05 = ? → × 1.05 = ?
$3,000 at 4% p.a. compounded annually for 2 years. Write down i, n and r, then find A. Check below.
When a child is born, a parent deposits $3,000 at 4.2% p.a. compounding monthly. Find the interest earned by the child's 18th birthday.
In one sentence, out loud: why does compound interest beat simple interest? If you can say it, you've got it.