Suppose you are renegotiating an allowance with your parents. Currently you are given $25 per week, but it is the first of June, and you have started mowing the lawn and taking out the trash every week, and you think your allowance should be increased.
Your father considers the situation and makes you the following offer:
"I tell you what, son. I will give you three options for your allowance, you tell me which you would like"
"Option A: You keep the $25 per week"
"Option B: You take $15 this week, then $16 next week, and so on. I'll continue adding $1 per week until New Year's."
"Option C: I'll give you 1 penny this week, and then double your allowance each week until the first of October, then keep it at that rate."
Which option would you choose?
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James Sousa: Compounded Interest
Guidance
Simple Interest vs. Compound Interest
Simple interest is interest which accrues based only on the principal of an investment or loan. The simple interest is calculated as a percent of the principal.
Simple Interest: \begin{align*}i = p \cdot r \cdot t\end{align*} .
Variable i is interest, p represents the principal amount, r represents the interest rate, and t represents the amount of time the interest has been accruing. For example, say you borrow $2,000 from a family member, and you insist on repaying with interest. You agree to pay 5% interest, and to pay the money back in 3 years.
The interest you will owe will be 2000(0.05)(3) = $300. This means that when you repay your loan, you will pay $2300. Note that the interest you pay after 3 years is not 5% of the original loan, but 15%, as you paid 5% of $2000 each year for 3 years.
Now let’s consider an example in which interest is compounded. Say that you invest $2000 in a bank account, and it earns 5% interest annually. How much is in the account after 3 years?
Compound interest: @$\begin{align*}A(t) = p \cdot (1 +r)^{t}\end{align*}@$
Here, A ( t ) is the A mount in the account after a given t ime in years, p rincipal is the initial investment, and r ate is the interest rate. Note that we use @$\begin{align*}(1 + r)\end{align*}@$ instead of just @$\begin{align*}r\end{align*}@$ , so we can find the entire amount in the account, not just the interest paid.
@$\begin{align*}A(t) = 2000 \cdot (1.05)^{3}\end{align*}@$
After three years, you will have $2315.25 in the account, which means that you will have earned $315.25 in interest.
Compounding results in more interest because the principal on which the interest is calculated increased each year. Another way to look at it is that compounding creates more interest because you are earning interest on interest, and not just on the principal.
Example A
Use the formula for compound interest to determine the amount of money in an investment after 20 years, if you invest $2000, and the interest rate is 5% compounded annually.
Solution:
The investment will be worth $5306.60
A(t) = P(1 + r) ^{ t }
A(20) = 2000(1.05) ^{ 20 }
A(20) = $5306.60
Example B
How long will it take for $2000, invested at 5% compounded annually, to reach $7,000?
Solution:
If we graph the function A(t) = 2000(1.05) ^{ t } , we can see the values for any number of years.
If you graph this function using a graphing calculator, you can determine the value of the investment by tracing along the function, or by pressing <TRACE> on your graphing calculator and then entering an x value.
You can also choose an investment value you would like to reach, and then determine the number of years it would take to reach that amount. Find the intersection of the exponential function with the line y = 7000.
You can see here that the line and the curve intersect at a little less than x = 26. Therefore it would take almost 26 years for the investment to reach $7000.
Example C
What is the value of an investment after 20 years, if you invest $2000, and the interest rate is 5% compounded continuously ?
Solution:
The more often interest is compounded, the more it increases, but there is a limit. Each time you increase the number of compoundings, you decrease the fraction of the annual interest that is applied to each compounding. Eventually, the differences become so small as to be negligible. This is known as continuous compounding .
The function A ( t ) = Pe ^{ rt } is the formula we use to calculate the amount of money when interest is continuously compounded, rather than interest that is compounded at discrete intervals, such as monthly or quarterly.

 A ( t ) = Pe ^{ rt }
 A (20) = 2000 e ^{ .05(20) }
 A (20) = 2000 e ^{ 1 }
 A (20) = $5436.56
Concept question wrapup: "Which option would you choose?" Assume you want to make the most money possible by the end of the year. Assume also that there are 24 weeks left. Option A = @$\begin{align*}25 \cdot 24 = \$600\end{align*}@$ total Option B = @$\begin{align*}15 + 16 + 17... + 39 = \$609\end{align*}@$ total Option C (assuming 16 weeks until Oct.) = @$\begin{align*}1 \cdot (2^{16}) = \$655.36\end{align*}@$ each week after Oct 1. It is entirely possible that dear old dad didn't take exponential growth seriously enough, he may need a second job! 

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Guided Practice
1) Compare the values of the investments shown in the table. If everything else is held constant, how does the compounding influence the value of the investment?

Principal r n t a. $4,000 .05 1 (annual) 8 b. $4,000 .05 4 (quarterly) 8 c. $4,000 .05 12 (monthly) 8 d. $4,000 .05 365 (daily) 8 e. $4,000 .05 8760 (hourly) 8
2) Determine the value of each investment.
 a. You invest $5000 in an account that gives 6% interest, compounded monthly. How much money do you have after 10 years?
 b. You invest $10,000 in an account that gives 2.5% interest, compounded quarterly. How much money do you have after 10 years?
3) How long will it take $2000 to grow to $25,000 at a 5% interest rate?
Answers
1) Use the compound interest formula. For this example, the n is the quantity that changes: @$\begin{align*}A(8) = 4000 \left (1 + \frac{.05} {n}\right )^{8n}\end{align*}@$

Principal r n t A a. $4,000 .05 1 (annual) 8 $5909.82 b. $4,000 .05 4 (quarterly) 8 $5952.52 c. $4,000 .05 12 (monthly) 8 $5962.34 d. $4,000 .05 365 (daily) 8 $5967.14 e. $4,000 .05 8760 (hourly) 8 $5967.29  A graph of the function @$\begin{align*}f(x) = 4000 \left (1 + \frac{.05} {x}\right )^{8x}\end{align*}@$ is shown below:
 The graph seems to indicate that the function has a horizontal asymptote at $6000. However, if we zoom in, we can see that the horizontal asymptote is closer to 5967.
 What does this mean? This means that for the investment of $4000, at 5% interest, for 8 years, compounding more and more frequently will never result in more than about $5968.00.
2)
 a. $5000, invested for 10 years at 6% interest, compounded monthly.

 @$\begin{align*}A(t) = P \left (1 + \frac{r} {n}\right )^{nt}\end{align*}@$
 @$\begin{align*}A(10) = 5000 \left (1 + \frac{.06} {12}\right )^{12\cdot 10}\end{align*}@$
 @$\begin{align*}A(10) = 5000 \left (1.005\right )^{120}\end{align*}@$
 @$\begin{align*}A(10) = \$9096.98\end{align*}@$
 b. $10000, invested for 10 years at 2.5% interest, compounded quarterly.

 Quarterly compounding means that interest is compounded four times per year. So in the equation, n = 4.

 @$\begin{align*}A(t) = P \left (1 + \frac{r} {n}\right )^{nt}\end{align*}@$
 @$\begin{align*}A(10) = 6000 \left( 1 + \frac{.025} {4}\right )^{4 \cdot 10}\end{align*}@$
 @$\begin{align*}A(10) = 6000 (1.00625)^{40}\end{align*}@$
 @$\begin{align*}A(10) = \$12,830.30\end{align*}@$
 In each example, the value of the investment after 10 years depends on three quantities: the principal of the investment, the number of compoundings per year, and the interest rate.
3) It will take about 50 years:

A ( t ) = Pe ^{ rt } 25,000 = 2000 e ^{ .05(t) } 12.5 = e ^{ .05(t) } Divide both sides by 2000 ln 12.5 = ln e ^{ .05( } ^{ t } ^{ ) } Take the ln of both sides ln 12.5 = .05 t ln e Use the power property of logs ln 12.5 = .05 t × 1 ln e = 1 ln 12.5 = 0.5 t Isolate t @$\begin{align*}t = \frac{ln 12.5} {.05} \approx 50.5\end{align*}@$
Explore More
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