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Apply the Annual Compound Interest Formula
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Simple interest only earns a fixed amount of interest based on the original principal amount. On the other hand, compound interest is calculated by taking the interest earned and adding it to the principal amount for the next interest earning period of time. Compound interest grows with each compounding period.

Here is a visual representation of the difference between simple interest and compound interest. Notice that in the simple interest images, the blocks representing interest always stay the same size. However, the blocks representing interest in the compound interest column get bigger with each period. This also makes the principal of the next period grow bigger as well.

We will start by just compounding one time per year, assuming that our rate is an annual rate.

Annual Compound Interest Formula:

$$A = P( 1 + r )^t$$

A = Amount (ending amount)

P = Principal (beginning amount)

r = Interest rate when compounding one time per year as a decimal

t = Time in years

Introduction to Compound Interest (3:56)

The next video uses simple interest to show an example of how compound interest works.

Introduction to Calculating Compound Interest

The following video will show another example of using simple interest to calculate compound interest, this time using excel to do those calculations. Then it will introduce the compound interest formula, which makes these calculations much faster.

Annual Compound Interest (8:13)

Compound interest adds the interest earned in the previous period to the principal amount, so the interest from previous periods also earns interest. This is an important tool to understand as it is used daily in loans, credit cards, and investments (including savings and some checking accounts). Spend the time you need to understand this principle because it will help you in your life.

### Practice Problems

Use the information given in Question 1 to solve Questions 2 and 3:

1. If you invest $$100$$ in an account that pays $$10\%$$ simple interest annually for one year, what is the total amount of money that you would have at the end of one year? (
Solution
Solution:
$$100 + 100 \times 10\% = 100 + 100 \times 0.10 = 100 + 10 = 110$$
)
2. Suppose that you take the total amount of money that you had after the first year in the previous problem and invest it in the same account for one more year. How much money would you have at the end of the second year? (
Solution
Solution:
$$110 + 110 \times 10\% = 110 + 110 \times 0.10 = 110 + 11 = 121$$
)
3. Suppose that you take the total amount of money that you had after the second year in the previous problem and invest it in the same account for yet another year. How much money would you have at the end of the third year? (
Solution
Solution:
$$121 + 121 \times 10\% = 121 + 121 \times 0.10 = 121 + 12.10 = 133.10$$
)

Use the annual compound interest formula for the following questions:

4. How much will you have at the end of three years if you invest $$100$$ in an account that pays $$10\%$$ annual interest compounded once a year? (You should get the same answer as in question 3.) (
Solution
Solution:
$$100(1 + 10\%)^3 = 100(1+0.1)3 = 100(1.1)^3 = 100(1.331) = 133.10$$
)
5. If you invest $$1200$$ in an account that pays $$8\%$$ interest compounded annually, what is the total amount of money that you would have at the end of three years? (
Solution
Solution:
$$1200(1+8\%)^3 = 1200(1+0.08)^3 = 1200(1.08)^3 = 1200(1.259712) = 1511.65$$
Details:
We are investing $$1200$$ that pays $$8\%$$ compounded annually for 3 years, so we will use the formula:

$${\text{A=P(1+r)}}^{\text{t}}$$

The first thing that we need to do is determine each of the following:

Principle = $${\color{MediumVioletRed}P}$$ = $${\color{MediumVioletRed}1200}$$

Rate = $${\color{DodgerBlue}r}$$ = $${\color{DodgerBlue}0.08}$$ (we must write it as a decimal for the formula)

Time = $${\color{MediumSeaGreen}t}$$ = $${\color{MediumSeaGreen}3}$$

Then we plug in each value into the formula:

$${\text{A=}{\color{MediumVioletRed}{\text{P}}}(1 + {\color{DodgerBlue}{\text{r}}})}^{\color{MediumSeaGreen}{\text{t}}}$$

$${\text{A=}{\color{MediumVioletRed}{\text{1200}}}(1 + {\color{DodgerBlue}{\text{0.08}}})}^{\color{MediumSeaGreen}{\text{3}}}$$

Add $$1$$ to $$0.08$$:

$${\text{A=}{\color{MediumVioletRed}{\text{1200}}}({\color{DodgerBlue}{\text{1.08}}})}^{\color{MediumSeaGreen}{\text{3}}}$$

Raise $$1.08$$ to the third power:

$${\text{A=}}{\color{MediumVioletRed}{\text{1200}}}({\color{DodgerBlue}{\text{1.259712}}})$$

$${\text{A=}}{\color{MediumVioletRed}{1511.6544}}$$

So we will have $$1511.65$$ at the end of 3 years.
)
6. If you invest $$1200$$ in an account that pays $$9\%$$ interest compounded annually, what is the total amount of money that you would have at the end of 25 years? (
Solution
Solution:
$$1200(1+9\%)^{25} = 1200(1+0.09)^{25}= 1200(1.09)^{25} = 1200(8.623081) = 10347.70$$
Details:
We are investing $$1200$$ that pays $$9\%$$ compounded annually for 25 years, so we will use the formula:

$${\text{A=P(1+r)}}^{\text{t}}$$

The first thing that we need to do is determine each of the following:

Principle = $${\color{MediumVioletRed}P}$$ = $${\color{MediumVioletRed}1200}$$

Rate = $${\color{DodgerBlue}r}$$ = $${\color{DodgerBlue}0.09}$$ (we must write it as a decimal for the formula)

Time = $${\color{MediumSeaGreen}t}$$ = $${\color{MediumSeaGreen}25}$$

Then we plug in each value into the formula:

$${\text{A=}{\color{MediumVioletRed}{\text{P}}}(1 + {\color{DodgerBlue}{\text{r}}})}^{\color{MediumSeaGreen}{\text{t}}}$$

$${\text{A=}{\color{MediumVioletRed}{\text{1200}}}(1 + {\color{DodgerBlue}{\text{0.09}}})}^{\color{MediumSeaGreen}{\text{25}}}$$

Add $$1$$ to $$0.09$$:

$${\text{A=}{\color{MediumVioletRed}{\text{1200}}}({\color{DodgerBlue}{\text{1.09}}})}^{\color{MediumSeaGreen}{\text{25}}}$$

Raise $$1.09$$ to the twenty-fifth power:

$${\text{A=}}{\color{MediumVioletRed}{\text{1200}}}({\color{DodgerBlue}{\text{8.62308}}})$$

Then multiply $$1200$$ by $$8.62308$$:

$${\text{A=}}{\color{MediumVioletRed}{10347.696792}}$$

So we will have $$10,347.70$$ at the end of 25 years.
)

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