Compound Interest Formula

Compound interest is interest that is calculated on both the money deposited and the interest earned from that deposit. The formula for compound interest is \(A=P(1+\frac{r}{n})^{nt}\), where \(A\) represents the final balance after the interest has been calculated for the time, \(t\), in years, on a principal amount, \(P\), at an annual interest rate, \(r\). The number of times in the year that the interest is compounded is \(n\).

Example 1:

Jasmine deposits $520 into a savings account that has a 3.5% interest rate compounded monthly. What will be the balance of Jasmine’s savings account after two years? To find the balance after two years, \(A\), we need to use the formula, \(A=P(1+\frac{r}{n})^{nt}\). The principal, \(P\), in this situation is the amount Jasmine used to start her account, $520. The rate, \(r\), as stated in the problem, is 3.5% (or 0.035 as a decimal) and compounded monthly, so \(n=12\). Since we are looking for the balance of the account after two years, 2, is the time, \(t\). \(A=520(1+\frac{0.035}{12})^{12(2)}\) \(A=557.65\) The balance of Jasmine’s account after 2 years is $557.65.

Example 2:

Lex has $1,780.80 in his savings account that he opened 6 years ago. His account has an annual interest rate of 6.8% compounded annually. How much money did Lex use to open his savings account? To find the principal, \(P\) we can use the same formula, \(A=P(1+\frac{r}{n})^{nt}\). We have the balance of the account, \(A\), after 6 years, which is $1,780.80. The interest rate, \(r\), is 6.8% (or 0.068 as a decimal) and is compounded annually, so \(n=1\). The time, \(t\), is 6, since we know he opened his account 6 years ago. Plug in the known values into the formula and solve for the missing variable, \(P\).

\(1{,}780.80=P(1+\frac{0.068}{1})^{1(6)}\) \(1{,}780.80=1.484P\) \(1{,}200=P\) The principal amount Lex used to open his account 6 years ago is $1,200.

Dominic and William each started the year by depositing $1,000 into savings accounts at their banks that each offer 4% interest rate. Even though neither of them made additional deposits, at the end of the year, Dominic had $1,040 in his account, while William had $1,040.60. How is this possible?

While Dominic and William both received a 4% interest rate, the fact that they use two different banks informs us that the interest on one of their savings accounts must compound more frequently than the other.

In fact, it turns out that Dominic’s account only collects simple interest, while William’s account has interest that compounds quarterly.

Calculating Simple Interest

Simple interest can be calculated using the formula \(i=Prt\), which states that the interest is equal to the principal times the annual interest rate times the number of years. Using this formula, we can find that Dominic collects $40 after one year, bringing his balance to $1,040.

\(i = Prt\)

\($40=$1,000\times 0.04\times 1\)

Calculating Compound Interest (Example 1)

But how did William end up with 60 extra cents? Well, we know that he still gets a 4% interest rate, but his interest is calculated and paid quarterly, every three months. And whatever interest he gets one quarter, he is able to gain interest on that interest in the following quarters.

This is what makes compound interest so interesting! Even if William never deposits anything else into his account, his interest will continue to grow larger and larger with each quarter. While slow at first, this forms exponential growth that becomes more and more powerful over time.

Balance After Quarter 1

Let’s work through the arithmetic of it. We know that William started the year with a principal of $1,000 and receives a 4% interest rate.

\(P = 1,000\)

\(r=0.04\)

Since his interest is compounded quarterly, he will receive a piece of that 4% interest every three months. At the end of the first quarter, William receives one-fourth of the 4%, which is 1%, applied to his current balance of $1,000.

Then, this first-quarter interest is added to his starting balance. So, we can represent his balance at the end of the first quarter as \(Q_1\) being equal to \(P+P(\frac{r}{4})\), or equivalently, \(P(1+\frac{r}{4})\).

\(Q_1=P(1+\dfrac{r}{4})=1,000(1+\dfrac{0.04}{4})\) \(=1,000(1+0.01)=1,010\)

So after the first quarter, William’s balance is $1,010.

Balance After Quarter 2

What will his balance be at the end of the second quarter?

To determine that amount, we want to apply the one-fourth of the 4% interest to William’s newest balance, \(Q_1\). So this time, we are going to have \(Q_2\), the balance at the end of the second quarter, equal to \(Q_1\) plus interest.

In other words, we can basically reuse the formula for \(Q_1\), but we are going to write \(Q_1\) in place of \(P\) now, because \(Q_1\) is the new “previous balance.”

\(Q_2=Q_1(1+\dfrac{r}{4})=1,000(1+0.01)\)

\(Q_2=1,010(1.01)\)

\(Q_2=1,020.10\)

This would give us a new balance of $1,020.10.

Before we move on to the third quarter balance, I want to point something out. When we wrote that \(Q_2=Q_1(1+\frac{r}{4})\), we were equivalently saying that \(Q_2=P(1+\frac{r}{4})(1+\frac{r}{4})\) because we already defined \(Q_1\) as being equal to \(P(1+\frac{r}{4})\).

This expression can be slightly condensed to \(Q_2=P(1+\frac{r}{4})^2\).

Balance After Quarter 3

All right, let’s look at the end of the third quarter. William’s balance at the end of the third quarter, \(Q_3,\) should then be equal to \(Q_2\) plus interest.

Since we just said that \(Q_2\) is equal to \(P(1+\frac{r}{4})^2\), we can use that in the calculation of \(Q_3\).

\(Q_3=Q_2(1+\dfrac{r}{4})\) \(=P(1+\dfrac{r}{4})^2(1+\dfrac{r}{4})\) \(=P(1+\dfrac{r}{4})^3\)

We now have that \(Q_3\) is equal to the initial balance, \(P\), times the quarterly interest cubed.

Plugging in William’s starting balance of $1,000 and the interest rate, \(r=0.04\), we get that \(Q_3=$1,030.30\).

In this quarter, the interest was greater than in the previous by an increment of ten cents.

Balance After Quarter 4

The fourth-quarter balance, \(Q_4\), can be calculated by taking \(Q_3\) and adding on the rest of the interest for the year.

\(Q_4=Q_3(1+\dfrac{r}{4})\) \(=P(1+\dfrac{r}{4})^3(1+\dfrac{r}{4})\) \(=P(1+\dfrac{r}{4})^4\)

Plugging in the numerical values for \(P\) and \(r\), we get that William’s ending balance is indeed $1,040.60.

In the final quarter, his interest grew again, and in the end he had 60 cents more than Dominic. While 60 cents is relatively inconsequential, that difference would grow exponentially from year to year! In fact, after ten years, the difference would be over $80, and after twenty years the difference would be over $400!

Through working that calculation, we saw that the balance at the end of each interest payment was equal to the starting principal times \(1+\frac{r}{4}\) raised to whatever power corresponded to that payment. We can generalize this and say that the ending balance, which we’ll call \(A\), after \(t\) years for an account with quarterly compounding interest is equal to \(P(1+\frac{r}{4})^{4t}\), since there are four payments per year.

But since some banks may compound interest only twice a year, or perhaps offer it twelve times a year, we can generalize this formula even further. If we let \(n\) equal the number of interest payments per year, then the ending balance of a compound interest savings account will be as follows:

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

Calculating Compound Interest (Example 2)

Let’s use the generalized formula in another example.

Marlise’s savings account has $500 at the beginning of the year. Her bank pays 5% interest annually, which is compounded daily. What is her balance after five years if she does not deposit any more money?

This is a bit of an extreme example, because it’s very unlikely to find a bank that pays 5% annual interest and compounds daily. Let’s work it out though and see what Marlise’s ending balance would be.

In this case, \(P=500\), \(r=0.05\), \(n=365\) (because the interest compounds daily), and \(t=5 \text{ years}\).

Plugging all of these into the compound interest formula, we get:

\(A=P(1+\frac{r}{n})^{nt}\) \(=500(1+\dfrac{0.05}{365})^{365\times 5}\)

With the help of a calculator, we can determine that this comes out to $642.00. That’s $142 in growth!

Calculating Compound Interest (Example 3)

Let’s work through one more example.

For the past 50 years, Mrs. Baker has been collecting interest on an initial deposit. Her savings account pays 3% interest annually, and compounds monthly. If her balance right now is $8,946.62, how much was her initial deposit?

This time, we know the ending balance, \(A\), but do not know the amount of the initial deposit, \(P\). We can solve for \(P\), though, by rearranging our compound interest formula.

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

Dividing both sides by \((1+\frac{r}{n})^{nt}\), we can get \(P\) all by itself.

\(P=\dfrac{A}{(1+\frac{r}{n})^{nt}}\)

Now, let’s plug in what we know. We know that Mrs. Baker has $8,946.62 as her ending amount, \(A\).

\(P=\dfrac{8,946.62}{(1+\frac{r}{n})^{nt}}\)

We also know that her annual interest rate r is equal to 3%, or 0.03, and that her interest compounds monthly, so \(n=12\).

\(P=\dfrac{8,946.62}{(1+\frac{0.03}{12})^{12(t)}}\)

Finally, we know that she’s been collecting this interest for 50 years, so \(t=50\).

\(P=\dfrac{8,946.62}{(1+\frac{0.03}{12})^{12(50)}}\)

With the help of a calculator, we can determine that all this is equal to 2,000, so Mrs. Baker started her account with $2,000 fifty years ago, and it has now grown to nearly five times that much.

\(P=2,000.00\)

Now that we’ve used the compound interest formula to find both the ending and starting balances of accounts, you may be unsurprised that it can also be used to solve for the annual interest rate, r, or t, the number of years passed!

While the compound interest formula is certainly more complex than the simple interest formula, you can commit it to memory with just a few minutes of dedicated focus, and the calculations are quick if you have access to a calculator.

Einstein is credited with saying that “compound interest is the most powerful force in the universe.” With a little practice, you’ll soon have this powerful force mastered!

Thanks for watching, and happy studying!

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