1The idea in one paragraph

Put money in a bank at 4% a year and, at the end of the year, the bank adds 4% of your balance. With compound interest that interest stays in the account, so next year you earn 4% on a bigger balance: interest earns interest. Each year the balance is multiplied by 1.04, which makes the balances a geometric sequence. If the bank adds interest more often, say every month, you multiply by a smaller factor more times, and the formula in the booklet handles both parts. Depreciation is the same idea running downwards: a car that loses 15% of its value each year is multiplied by 0.85 each year. And inflation means that even a growing balance may buy less than it seems, so you sometimes need its real value, measured in today's prices.

2Simple interest first

Simple interest is paid only on the original amount. €5,000 at 4% simple interest earns 4% of €5,000, which is €200, every single year. After n years the balance is 5000 + 200n: an arithmetic sequence (1.2).

Compound interest is paid on the balance at the time, which includes all the earlier interest. After one year €5,000 becomes 5000 × 1.04 = €5,200, the same as simple interest. In the second year, though, the 4% is taken of €5,200, so the interest is €208, not €200. The gap is small at first and then grows every year. Figure 1 shows 20 years of each.

After 20 years simple interest has produced €9,000 and compound interest €10,955.62. The straight line is arithmetic; the curve bending away is geometric. That is 1.3's comparison, with money.

3Compound interest as a geometric sequence

Compounded once a year at r%, the balance is multiplied by (1 + r/100) each year:

Here PV, the present value, is the amount invested at the start, and FV, the future value, is what it grows to. The exponent is the number of times interest has been added, which is the number of years that have passed.

€5,000 at 4% a year, compounded yearly, for 6 years:

Notice the link with 1.3. If you set u₁ = PV, then the balance after n years is uₙ₊₁, because the first term is the balance after zero years. That is why this subtopic writes the formula with PV and n years rather than with u₁ and n terms: it avoids the one-year slip.

4Compounding more than once a year

Banks often add interest half-yearly, quarterly or monthly. The rate they quote is the nominal annual rate, r%. Compounding k times a year means:

• each period's rate is r% ÷ k, so each period multiplies by 1 + r/(100k);
• in n years there are kn periods.

FV = PV × (1 + r/100k)ᵏⁿ (in the formula booklet)

Figure 2 labels every part, because the marks are in reading r, k and n correctly from the words of the question.

The same €5,000 at a nominal 4% for 6 years, compounded monthly:

Figure 3 puts the four standard choices of k side by side.

More frequent compounding always gives a little more, because interest starts earning interest sooner. But the extra gets smaller each time: going from yearly to half-yearly adds €14.61, from quarterly to monthly only €5.04.

Two traps live in the formula. The first is to use 4% per month when the question says 4% a year compounded monthly: r/100k, not r/100. The second is to raise to the power n instead of kn: 72 periods, not 6.

5The GDC finance solver

The GDCs used in IB examinations have a finance solver (often called TVM, for "time value of money"). It is the same formula with the unknown made into a button, and on Paper 2 it is usually the quickest route. The guide says exam questions may require it. Figure 4 shows it filled in for the monthly example.

The sign convention is the solver's way of tracking which way money flows. Money you hand over (the investment) is negative; money you get back is positive. If PV and FV have the same sign, the solver will report an error or a nonsense answer.

Because the solver can find any one field from the others, it answers four kinds of question. In your written answer, list the values you entered: that is your method.

Find the future value. Shown above: N = 72, I% = 4, PV = −5000, PMT = 0, P/Y = C/Y = 12, solve for FV = €6,353.71.

Find the time. How long does €5,000 take to grow to at least €8,000 at a nominal 4.5% a year, compounded monthly?

The balance first reaches €8,000 after 126 months, 10 years and 6 months. If the question asks for the number of whole years, it is 11: after 10 whole years the balance is €7,834.96, which is not enough. Always round a time up to the next complete period, and check the two values either side if in doubt.

Find the rate. What nominal annual rate, compounded quarterly, turns €2,000 into €2,600 in 5 years?

Find the present value. How much must be invested now at 3% a year, compounded half-yearly, to have €10,000 in 8 years?

The minus sign is the solver telling you this is money paid in. Write the answer as a positive amount in words.

You can always check a solver answer with the formula: 2000 × (1 + 5.2818/400)²⁰ = 2600.00. ✓ The guide also says you will not be asked to derive the formula in the exam, so your effort belongs in using it.

6Depreciation

Depreciation is the loss in value of an asset over time, such as a car, a phone or a machine. In this course it is annual depreciation at a fixed percentage: each year the value is multiplied by (1 − r/100). It is compound interest with the rate subtracted.

value after n years = initial value × (1 − r/100)ⁿ

With the solver, enter I% as a negative number (−15 for 15% depreciation), with P/Y = C/Y = 1.

A car is bought for €24,000 and depreciates by 15% a year. Figure 5 follows its value.

When does it first fall below €8,000? Solve 24 000 × 0.85ⁿ < 8000 with the solver (FV = 8000, solve for N) or with a table of values:

The value first falls below €8,000 after 7 years. Again: round up to a whole year, and show the years either side.

The loss in value is initial value minus current value: after 5 years the car has lost 24 000 − 10 648.93 = €13,351.07. Read carefully whether the question wants the value or the loss.

7Inflation and real value

Inflation is the rise in the general level of prices. If inflation is 2.5% a year, something that costs €100 today costs €102.50 next year. So €102.50 next year buys only what €100 buys today.

That means an investment can grow in euros while growing less, or not at all, in what it will buy. The nominal value is the number of euros in the account. The real value is what those euros are worth in today's prices: their buying power.

The standard method in the exam is to use a real interest rate:

real interest rate ≈ nominal interest rate − inflation rate

and put that rate into the compound interest formula or the solver. The answer is the real value in today's money.

€10,000 is invested at 5% a year, compounded yearly, for 8 years. Inflation is 2.5% a year.

Figure 6 shows the two curves and the gap between them.

The account will hold €14,774.55, but that money will buy only what €12,184.03 buys today. The shaded gap is the buying power lost to inflation. If inflation were higher than the interest rate, the real rate would be negative and the real value would fall, even though the balance rises.

The subtraction is an approximation. A more exact method divides the nominal value by the growth in prices, 14 774.55 ÷ 1.025⁸ = €12,126.16. The two differ by a few tens of euros here. Use the method the question describes; if it says "use the real interest rate", subtract the rates, and state the rate you used either way so the examiner can follow you.

8Beyond the syllabus: where e comes from

This is enrichment and it is not examined. Suppose a nominal 100% a year is compounded k times a year, so €1 becomes (1 + 1/k)ᵏ after one year:

As k grows, the value creeps up but never passes a limit. That limit is the number e ≈ 2.718 28, which you meet as the base of the natural logarithm in 1.5 and in exponential functions in Topic 2. Compounding "continuously" would multiply by e each year.