Compound Interest

Learning goals

Contrast linear simple interest with exponential compounding
Apply $A = P(1+r)^t$ to find a balance or a needed principal
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. Suppose you deposit 1000 dollars at $5\%$ simple interest. Every year the account earns $5\%$ of the original 1000 dollars, which is $1000 \times 0.05 = 50$ dollars, whether it is the first year or the tenth. After $3$ years the interest is $3 \times 50 = 150$ dollars, so the balance is $1150$ dollars.

Now write that so it works for any deposit and rate. 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 $r = 0.05$ . 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

Many savings accounts do not 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, and that 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. Equal steps in $t$ multiply the balance by the constant $1 + r$ , which is exactly the pattern of an exponential function from the previous lesson: $f(t) = P(1+r)^t$ with base $1 + r$ . Since $r > 0$ makes that base greater than $1$ , 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 $800$ dollars in an account paying $4\%$ compounded annually. What is the balance after $3$ years?

Convert the rate to a decimal, $r = 0.04$ , and read off the pieces of the formula: $P = 800$ , $r = 0.04$ , and $t = 3$ . Since compounding is annual, apply $A = P(1+r)^t$ directly:

A = 800\,(1 + 0.04)^3 = 800\,(1.04)^3.

Now raise the factor to the third power. Squaring gives $1.04^2 = 1.0816$ , and one more factor gives $1.04^3 = 1.124864$ , so

A = 800 \times 1.124864 = 899.8912.

Rounded to the nearest cent, the balance is $899.89$ dollars. Simple interest would have given only $800(1 + 0.04 \times 3) = 896$ dollars, so compounding is worth an extra $3.89$ 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} \approx 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.

Check your understanding

You want $3000$ dollars in $2$ years at $8\%$ compounded annually. Using $1.08^2 = 1.1664$ , about how much must you deposit now?

3499.20

dollars B

2586.21

dollars C

2572.02

dollars D

2520.00

dollars

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.

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?

Compounding quarterly splits the year into $4$ periods. Each period pays only one quarter of the yearly rate, $\dfrac{0.08}{4} = 0.02$ , and over $1$ year there are $4$ of those periods, so the factor $1.02$ is applied $4$ times:

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 \approx 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.

Two things changed in that calculation, and both are worth naming so the idea works for any account. Compounding $n$ times a year means each period is only a fraction of a year, so it pays only a fraction of the yearly rate, namely $\dfrac{r}{n}$ (here $\dfrac{0.08}{4}$ ). And over $t$ years there are $nt$ periods in total (here $4 \times 1 = 4$ ), not $t$ . 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 $n$ . Quarterly compounding for $3$ years, for instance, means $n = 4$ and $t = 3$ , so the exponent is $4 \times 3 = 12$ , not $3$ .

Check your understanding

A sum is compounded monthly at $6\%$ for $3$ years. Which expression gives the balance $A$ ?

P(1+0.06)^{36}

P(1+0.005)^{3}

P(1+0.005)^{36}

P(1+0.06)^{3}

Worked example 4 Transferring the method: monthly compounding

You invest $1000$ dollars at $12\%$ compounded monthly for $2$ years. What is the balance?

Monthly means $n = 12$ , so each month pays $\dfrac{0.12}{12} = 0.01$ , and over $2$ years there are $nt = 12 \times 2 = 24$ periods. Substitute into the formula:

A = 1000\left(1 + \frac{0.12}{12}\right)^{12\cdot 2} = 1000\,(1.01)^{24}.

Raising $1.01$ to the $24$ th power needs a calculator’s power key: $1.01^{24} \approx 1.26973465$ . Keep those digits until the last step, so

A \approx 1000 \times 1.26973465 = 1269.73465.

The balance is about $1269.73$ dollars. The steps are exactly the ones from the quarterly example, just with a different $n$ and $t$ .

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. Choosing a rate of $100\%$ , so $r = 1$ , makes the compounding factor exactly $\left(1 + \dfrac{1}{m}\right)^{m}$ , the pattern that defines $e$ . 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 \approx 2.71828\ldots

This 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. The same idea holds when $n$ in $A = P\left(1 + \frac{r}{n}\right)^{nt}$ grows without bound: the growth factor approaches $e^{rt}$ , the same limit as above but built from a general rate $r$ instead of $r = 1$ . Restoring the principal, the balance 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. The five schedules are ordered left to right by frequency, spaced evenly for readability rather than by their numeric value.

Worked example 5 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.

Quarterly and monthly follow the same steps as the earlier worked examples, just with $r = 0.06$ :

Schedule	Rate per period, $r/n$	Periods, $nt$	Balance
Quarterly	$0.015$	$4$	$1061.36$
Monthly	$0.005$	$12$	$1061.68$

Compounded continuously, use $A = Pe^{rt}$ with $e^{0.06} \approx 1.061837$ (a calculator’s $e^x$ key gives this directly):

A = 1000\,e^{0.06} \approx 1000 \times 1.061837 \approx 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 next question is how long an investment takes to double, which means solving $(1 + r)^t = 2$ for the exponent $t$ . That needs a new tool, the logarithm, covered in the next lesson; every quantity solved for in this lesson is a balance $A$ or a principal $P$ , never an exponent.

Common mistakes

Using the percent instead of the decimal. A rate of $5\%$ is $r = 0.05$ , 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 $nt$ . 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, $A = P(1 + rt)$ , grows in a straight line. Compound interest, $A = P(1+r)^t$ , curves upward because it pays interest on interest. They agree after one period and then separate.
Thinking more frequent compounding grows without bound. The gains shrink and level off at 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 an intermediate balance first can throw the answer off by several cents.

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, $A = Pe^{rt}$ , the ceiling every finite schedule climbs toward but never passes.

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)

Core practice