Compound Interest

Learning goals

Contrast linear simple interest with exponential compounding
Apply $A = P(1 + r)^t$ for yearly compounding
Divide the rate by $n$ and multiply the exponent by $n$
Convert a percent to its decimal rate first
Reach $A = Pe^{rt}$ by letting the periods grow without bound

Simple interest: the same amount every year

The plainest way to pay interest is to base it only on the original deposit, called the principal. If a principal $P$ earns a yearly rate $r$ , then each year it earns the same fixed amount $Pr$ . This is simple interest. After $t$ years the total interest is

I = P r t,

and the balance is the principal plus that interest,

A = P + Prt = P(1 + rt).

The rate $r$ is always the decimal form of the percent, so a rate of $5\%$ means $r = 0.05$ . Suppose you deposit 1000 dollars at $5\%$ simple interest. Every year it earns $1000 \times 0.05 = 50$ dollars, the same 50 whether it is the first year or the tenth. After $3$ years the interest is $1000 \times 0.05 \times 3 = 150$ dollars, and the balance is $1150$ dollars. Because the yearly increase never changes, simple interest grows in a straight line: equal steps in time add equal amounts of money.

Compound interest: interest on its own interest

Real accounts almost never work that way. At the end of each period the interest is added to the balance, and the next period’s interest is figured on that larger balance. The interest starts earning interest. This is compound interest, and it changes the arithmetic from adding to multiplying.

Track the same 1000 dollars at $5\%$ , but now compounded once a year. After the first year the balance is the principal plus $5\%$ of it, which is $1000 + 50 = 1050$ . For the second year the $5\%$ applies to $1050$ , not to the original $1000$ , so the account earns $1050 \times 0.05 = 52.50$ and the balance becomes $1102.50$ . The extra $2.50$ beyond simple interest’s flat $50$ is interest earned on the first year’s interest. A third year multiplies again, and the balance reaches $1157.625$ , which rounds to $1157.63$ dollars. Compare the two methods after three years: simple interest gives $1150$ dollars, compound gives $1157.63$ dollars, and the gap only widens as the years pile up.

The key move is that each year multiplies the balance by the same factor. Adding $5\%$ to a balance is the same as multiplying it by $1.05$ , because $B + 0.05B = 1.05B$ . That single observation turns the year-by-year story into a clean formula.

Why annual compounding gives $A = P(1+r)^t$ #

Start with a principal $P$ and an annual rate $r$ , compounded once a year. During the first year the account earns interest $Pr$ , so the balance climbs to $P + Pr$ . Factor out the common $P$ : one year multiplies the balance by $1 + r$ , leaving $P(1+r)$ .

Nothing about the second year is different except where it begins. Its starting balance is $P(1+r)$ , and one more year multiplies that by the same factor $1 + r$ , so the balance becomes $P(1+r)(1+r) = P(1+r)^2$ . The third year repeats the move exactly, giving $P(1+r)^3$ , and after $t$ whole years the factor $1 + r$ has been applied $t$ times in a row:

A = P(1 + r)^t.

The exponent counts the years precisely because each year contributes one more factor of $1 + r$ , no more and no less. This is the pattern you met in the previous lesson. Equal steps in $t$ multiply the output by the constant $1 + r$ , which is the signature of an exponential function. The balance is the exponential function $f(t) = P(1+r)^t$ with base $1 + r$ . Since $r > 0$ makes the base $1 + r > 1$ , it is exponential growth, which is why a compounded balance eventually outruns any straight line.

The same 1000 dollars at 10 percent for 10 years. Simple interest adds a flat 100 dollars a year, so it climbs in a straight line to 2000. Compound interest multiplies by 1.10 each year, so it curves upward, reaching about 2594. The two start together and then the curve pulls away, because compound interest earns interest on its own interest.

Worked example 1 A balance compounded annually

You deposit $1000$ dollars in an account paying $5\%$ compounded annually. What is the balance after $3$ years?

Convert the rate to a decimal, $r = 0.05$ , and read off the pieces of the formula: $P = 1000$ , $r = 0.05$ , and $t = 3$ . Since compounding is annual, apply $A = P(1+r)^t$ directly:

A = 1000\,(1 + 0.05)^3 = 1000\,(1.05)^3.

Now raise the factor to the third power. Squaring gives $1.05^2 = 1.1025$ , and one more factor gives $1.05^3 = 1.157625$ , so

A = 1000 \times 1.157625 = 1157.625.

Rounded to the nearest cent, the balance is $1157.63$ dollars. Simple interest would have given only $1000(1 + 0.05 \times 3) = 1150$ dollars, so compounding is worth an extra $7.63$ dollars over these three years, all of it interest earned on earlier interest.

Check your understanding

You invest $500$ dollars at $10\%$ compounded annually. What is the balance after $2$ years?

550

dollars B

600

dollars C

605

dollars D

610

dollars

Because the formula is just multiplication, you can also run it backward to find the deposit needed to reach a goal. If $A = P(1+r)^t$ , then dividing both sides by $(1+r)^t$ isolates the principal:

P = \frac{A}{(1+r)^t}.

Worked example 2 Working out the deposit you need

How much must you deposit today, at $6\%$ compounded annually, to have exactly $5000$ dollars in $2$ years?

Here the unknown is the principal $P$ , while $A = 5000$ , $r = 0.06$ , and $t = 2$ are known. Rearrange the formula to solve for $P$ by dividing:

P = \frac{A}{(1+r)^t} = \frac{5000}{(1.06)^2}.

Compute the denominator first, $1.06^2 = 1.1236$ , and then divide:

P = \frac{5000}{1.1236} = 4449.98.

You need to deposit about $4449.98$ dollars now. Notice that no logarithm was needed: the unknown was a plain factor, not the exponent, so ordinary division finished the job. Pulling an unknown out of the exponent is a different task, and it waits for the next lesson.

Compounding several times a year

Most accounts do not wait a whole year to compound. They might do it every six months, every quarter, or every month. Two things change when interest is compounded $n$ times a year. Each period is a fraction of a year, so it pays only a fraction of the yearly rate, namely $\dfrac{r}{n}$ . And over $t$ years there are not $t$ periods but $n$ periods per year for $t$ years, a total of $nt$ periods. Each period still multiplies the balance by “one plus its own rate,” and there are $nt$ of them, so the factor $\left(1 + \dfrac{r}{n}\right)$ is applied $nt$ times:

A = P\left(1 + \frac{r}{n}\right)^{nt}.

Read the two edits carefully, because both are easy to drop. The rate inside the parentheses is divided by $n$ , and the exponent is multiplied by $n$ . The exponent is $nt$ , not $t$ . Quarterly compounding for $3$ years, for instance, means $n = 4$ and $t = 3$ , so the exponent is $4 \times 3 = 12$ , not $3$ .

Worked example 3 Quarterly compounding

You invest $1000$ dollars at $8\%$ compounded quarterly for $1$ year. What is the balance, and how does it compare with annual compounding?

Quarterly means $n = 4$ , so each quarter pays a rate of $\dfrac{0.08}{4} = 0.02$ , and over $1$ year there are $nt = 4 \times 1 = 4$ periods. Substitute into the formula:

A = 1000\left(1 + \frac{0.08}{4}\right)^{4\cdot 1} = 1000\,(1.02)^4.

Raising $1.02$ to the fourth power gives $1.02^4 = 1.08243216$ , so

A = 1000 \times 1.08243216 = 1082.43.

The balance is about $1082.43$ dollars. Compounding just once a year would have given $1000(1.08) = 1080$ dollars, so splitting the same $8\%$ into four quarterly steps earns an extra $2.43$ dollars. More frequent compounding pays a little more, because interest starts earning interest sooner.

Check your understanding

A sum is compounded quarterly for $5$ years. In the formula $A = P\left(1 + \frac{r}{n}\right)^{nt}$ , what number goes in the exponent $nt$ ?

9

20

5

4

Squeezing the periods: continuous compounding and the number e

If compounding four times a year beats once a year, what about a hundred times, or a million? Push the idea to its limit and compound not just every day but every instant. You might expect a balance compounded infinitely often to blow up to infinity, but it does not. It climbs toward a definite ceiling, and that ceiling introduces one of the most important numbers in mathematics.

To see the ceiling with clean numbers, strip away the principal and rate and ask what happens to the factor $\left(1 + \dfrac{1}{m}\right)^{m}$ as $m$ grows. This is exactly the compounding factor for a rate of $100\%$ compounded $m$ times, the sharpest case. Here is what the arithmetic gives:

$m$	$\left(1 + \tfrac{1}{m}\right)^{m}$
$1$	$2$
$2$	$2.25$
$5$	$2.48832$
$10$	$2.59374$
$100$	$2.70481$
$1{,}000$	$2.71692$
$10{,}000$	$2.71815$

The values keep rising, but by smaller and smaller amounts, and they crowd toward a single number near $2.71828$ . That limiting value is named $e$ :

e = \lim_{m \to \infty}\left(1 + \frac{1}{m}\right)^{m} \approx 2.71828\ldots

Like $\pi$ , the number $e$ is irrational: its decimals run forever without repeating. It is the base of the exponential function $e^x$ mentioned at the end of the previous lesson, and compound interest is where it is born.

Now bring the principal and rate back and let $n$ grow without bound in $A = P\left(1 + \frac{r}{n}\right)^{nt}$ . By the power-of-a-power rule, the growth factor $\left(1 + \frac{r}{n}\right)^{nt}$ factors as

\left(1 + \frac{r}{n}\right)^{nt} = \left[\left(1 + \frac{r}{n}\right)^{n}\right]^{t}.

Set $m = \dfrac{n}{r}$ , so that $\dfrac{r}{n} = \dfrac{1}{m}$ and $n = mr$ . The inner bracket becomes

\left(1 + \frac{1}{m}\right)^{mr} = \left[\left(1 + \frac{1}{m}\right)^{m}\right]^{r}.

As $n$ grows without bound so does $m$ , and the inner piece $\left(1 + \frac{1}{m}\right)^{m}$ marches toward $e$ . So the bracket approaches $e^{r}$ , and the whole growth factor approaches $\left(e^{r}\right)^{t} = e^{rt}$ . Restoring the principal, the balance $A = P\left(1 + \frac{r}{n}\right)^{nt}$ approaches $Pe^{rt}$ . This limiting case is continuous compounding, and its formula is

A = P\,e^{rt}.

One year on 1000 dollars at 6 percent, compounded more and more often. Annual compounding gives 1060.00; quarterly, monthly, and daily each earn a bit more, but the gains shrink and the balances press up against the continuous ceiling of 1061.84 dollars. Compounding more often never breaks past that dashed line.

Worked example 4 Annual, quarterly, monthly, and continuous side by side

Take $1000$ dollars at $6\%$ for $1$ year and compare four compounding schedules. Each uses the same $P = 1000$ , $r = 0.06$ , and $t = 1$ ; only the frequency $n$ changes.

Compounded annually, $n = 1$ :

A = 1000\,(1.06)^1 = 1060.00.

Compounded quarterly, $n = 4$ , so each period pays $\tfrac{0.06}{4} = 0.015$ and the exponent is $4$ :

A = 1000\,(1.015)^4 = 1000 \times 1.06136 = 1061.36.

Compounded monthly, $n = 12$ , so each period pays $0.005$ and the exponent is $12$ :

A = 1000\,(1.005)^{12} = 1000 \times 1.06168 = 1061.68.

Compounded continuously, use $A = Pe^{rt}$ with $e^{0.06} \approx 1.061837$ :

A = 1000\,e^{0.06} = 1000 \times 1.061837 = 1061.84.

Line them up: $1060.00 < 1061.36 < 1061.68 < 1061.84$ . Each jump in frequency adds a bit more, but the additions shrink, and every finite schedule stays below the continuous value $1061.84$ dollars. That continuous amount is the ceiling all the others are climbing toward.

Check your understanding

As the number of periods $m$ grows without bound, the factor $\left(1 + \frac{1}{m}\right)^{m}$ gets closer and closer to which value?

1

2

e \approx 2.718

D it grows without bound

A natural question is how long an investment takes to double. For annual compounding that means solving $(1 + r)^t = 2$ for $t$ , an unknown sitting in the exponent. You can bracket the answer by running the formula forward and watching for the balance to cross twice the principal. Solving for the exponent exactly needs a new tool. That tool is the logarithm, the subject of the next lesson. In this lesson every quantity you solve for is either the balance $A$ or the principal $P$ . Both are found by running the formula forward or dividing, never by undoing an exponent.

Common mistakes

Using the percent instead of the decimal. A rate of $5\%$ is $r = 0.05$ in the formula, not $5$ . Writing $A = 1000(1 + 5)^t$ multiplies the balance by $6$ every year. Always divide the percent by $100$ first: $5\% \to 0.05$ , $8\% \to 0.08$ .
Forgetting the exponent is $nt$ . When interest compounds $n$ times a year, the exponent is the total number of periods $nt$ , not the number of years $t$ . Quarterly for $3$ years is $\left(1 + \frac{r}{4}\right)^{12}$ , not $\left(1 + \frac{r}{4}\right)^{3}$ . Divide the rate by $n$ and multiply the exponent by $n$ .
Confusing simple and compound interest. Simple interest uses $A = P(1 + rt)$ and grows in a straight line, adding the same amount each year. Compound interest uses $A = P(1+r)^t$ and curves upward, because it pays interest on interest. They agree after one period and then separate.
Thinking more frequent compounding grows without bound. Compounding more often does earn more, but the gains shrink and level off. The balance never passes the continuous ceiling $A = Pe^{rt}$ . Infinitely frequent compounding gives a finite answer, not an infinite one.
Rounding too early. Round only the final balance to the nearest cent. Rounding the growth factor or an intermediate balance first can throw the answer off by several cents over many periods.

Key takeaways

Simple interest is linear: $A = P(1 + rt)$ adds the same amount $Pr$ every year.
Compound interest is exponential: reinvesting makes each period multiply the balance by $1 + r$ , so after $t$ years $A = P(1 + r)^t$ , the exponential function with base $1 + r$ .
Compounding $n$ times a year pays rate $\frac{r}{n}$ per period over $nt$ periods, giving $A = P\left(1 + \frac{r}{n}\right)^{nt}$ . The exponent is $nt$ , not $t$ .
The rate $r$ is always the decimal form of the percent ( $5\% \to 0.05$ ), and solving for the principal is just $P = \dfrac{A}{(1+r)^t}$ .
Letting $n$ grow without bound gives continuous compounding, $A = Pe^{rt}$ , where $e = \lim_{m\to\infty}\left(1 + \frac{1}{m}\right)^{m} \approx 2.71828$ . Finite schedules climb toward this ceiling but never pass it.

Compound Interest

Learning goals

Simple interest: the same amount every year

Compound interest: interest on its own interest

Compounding several times a year

Squeezing the periods: continuous compounding and the number e

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)