Compounding and Continuous Compounding of Interest

What is simple interest and interest compounding? Full explanations with examples and easy-to-use online calculators are presented to help you experiment with different parameter values and understand the formulas related to interest compounding. Problems on compound interest with detailed solutions are also included.

What Is Percent Increase?

To understand compounding, you first need to understand the percentage increase of a quantity. If $P$ is a quantity increased by a percentage rate $r$, then the new quantity is:

\[ P + rP = P(1 + r) \]

You need to retain the following key idea:

\[ \text{A quantity } P \text{ increased by a rate } r \text{ becomes } P(1+r) \]

Example 1.

$200$ increased by $5\%$ becomes:

\[ 200 + \frac{5}{100}\cdot 200 = 200(1+0.05) \]

Simple Interest (Not Compounded)

If an amount $P$ is deposited in a savings account at an interest rate $r$ that is not compounded, then after $t$ years the interest earned $I$ is:

\[ I = Prt \]

The total amount $A$ in the account after $t$ years is:

\[ A = P + Prt = P(1 + rt) \]

Example.

$ \$100 $ invested at a $ 3\% $ interest rate (not compounded) for 5 years earns:

\[ 100 \cdot \frac{3}{100} \cdot 5 = 15 \]

Yearly Interest Compounding

An amount of money $P$ (principal) is invested at an annual interest rate $r$.

At $t=0$: $A=P$
At $t=1$: \[ A = P + rP = P(1+r) \]
At $t=2$: \[ A = P(1+r) + rP(1+r) = P(1+r)^2 \]
At $t=3$: \[ A = P(1+r)^3 \]
At $t=4$: \[ A = P(1+r)^4 \]

By extension, after $t$ years:

\[ A = P(1+r)^t \]

So if an amount $P$ is invested at an annual rate $r$ and compounded yearly, the total amount after $t$ years is:

\[ A = P(1+r)^t \]

Example 2.

$ \$1000 $ invested for 3 years at $ 3\% $ compounded yearly:

\[ A = 1000(1+0.03)^3 = 1092.73 \]

Interest Compounded n Times Per Year

If the interest is compounded $n$ times per year, the amount after $t$ years is:

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

Example 3.

$ \$1000 $ invested for 3 years at $3\%$, compounded twice per year:

\[ A = 1000\left(1+\frac{0.03}{2}\right)^6 = 1093.44 \]

Continuous Compounding

Starting from:

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

As $n \to \infty$, the expression $\left(1 + \frac{1}{N}\right)^N$ approaches the constant $e \approx 2.71828$.

Hence, for continuous compounding:

\[ A = Pe^{rt} \]

Example 4.

$ \$1000 $ invested for 3 years at $3\%$, compounded continuously:

\[ A = 1000e^{0.03 \cdot 3} = 1094.17 \]

Conclusion.

Compare how the same $ \$1000 $ grows under simple interest, annual compounding, and continuous compounding, and determine which method earns the most.