## Simple and Compound Interest

Many resources show simple interest as i = Prt and compound interest as A = P(1 + i)n.
This is somewhat confusing since i is a dollar value in the first and interest rate per compounding period in the second.

Consider using the amount with compound interest found in other resources: A = P(1 + r/n)nt

A = Amount, P = Principle, r = interest rate (%), n = number of compounding periods per time interval, t = time

## Modelling Compound Interest, A = P(1 + i)n

• P = principal
• i = interest rate per compounding period (rate per number of compounding periods per time interval)
• n = number of compounding periods (compounding period * time interval)

Principal (P) = \$1000, Annual interest rate(r) = 5%, compounding periods(n)= 7 years

Step 1: Define the functions required for each of the four columns.

 Year Number Opening Balance Interest Earned Closing Balance 1 1000 50 1000(1.05) 2 1000(1.05) 50(1.05)1 1000(1.05)2 3 1000(1.05)2 50(1.05)2 1000(1.05)3 4 1000(1.05)3 50(1.05)3 1000(1.05)4 5 1000(1.05)4 50(1.05)4 1000(1.05)5 6 1000(1.05)5 50(1.05)5 1000(1.05)6 7 1000(1.05)6 50(1.05)6 1000(1.05)7 General Term 1000(1.05)n - 1 50(1.05)n - 1 1000(1.05)n

The opening balance for each year can be represented by the geometric sequence 1000, 1000(1.05), 1000(1.05)2, …;
The interest earned each year can be represented by the geometric sequence 50, 50(1.05)1, 50(1.05)2, …
The closing balance for each year can be represented by the geometric sequence 1000(1.05), 1000(1.05)2, 1000(1.05)3, …

Step 2:

 A B C D E 1 Year Opening Balance Interest Rate Interest Earned Closing Balance 2 1 \$1000 5.00% =B2*C2 =B2+D2 3 =A2+1 =E2 5.00% =B3*C3 =B3+D3 4 =A3+1 =E3 5.00% =B4*C4 =B4+D4 5 =A4+1 =E4 5.00% =B5*C5 =B5+D5 6 =A52+1 =E5 5.00% =B6*C6 =B6+D6 7 =A6+1 =E6 5.00% =B7*C7 =B7+D7 ... Continue for rest of spreadsheet

Graphing Calculator

Step 2: Because you are working with sequences and money, set your graphing calculator to 2 decimal places and in Seq mode.

Step 3: Determine the general term of each sequence.

Step 4: Enter the general term for the Opening Balance column as u(n).

Step 5: Enter the general term for the Interest column as v(n).

Step 6: Enter the general term for the Closing Balance column as w(n).

Image Club Studiogear/Getty Images

Step 7: Press to obtain the table.

Note: Use the arrow keys to scroll to the third commn, w(n).

 Example 1 Gina has \$3000 to invest. If she is quoted an interest rate of 10%/a, compounded quarterly, calculate the amount of Gina’s investment after 2 years. Solution Calculate the interest rate per period. Calculate the number of compounding periods. Example 2 Anna deposits \$100 in an account bearing interest at 10%, compounded annually. The interest is deposited into her account at the end of each year. Graph the amount of money she has in her account over a 5-year period. Also, graph the exponential function you used to calculate the accumulated amount on your graphing calculator. Use your graphs to determine when her original investment will grow by 50%. Solution Use the compound interest formula, A(t) = P(1 + i)n, to determine the amount in the account at the end of each year. P = 100, i = 10% = 0.10, and n = t A(t) = 100(1 + 0.10)t             = 100(1.1)t An exponential function for the amount of the investment after t years is A(t) = 100 (1.1)t. Graph these values. Remember: The amount is constant until the interest is added. That means the graph is discontinuous. This is an example of a step function—a function whose graph increases in discrete amounts and is constant between increments. Now, graph the function A(t) = 100(1.1)t on your graphing calculator using the window settings shown. [ y= ] [ 1 ] [ 0 ] [ 0 ] [ X ] [ 1 ] [ · ] [ 0 ] [ ^ ] [ X, T,, n] [ GRAPH ]   When Anna’s investment grows by 50%, she will have \$100 +50% of \$100 = \$100 + \$50           = \$150 If you look at the first graph, Anna never has exactly \$150.00. She will have to wait 5 years for her investment to grow by at least 50%. At the end of the fifth year, she will have \$161.05. To find this answer using your graphing calculator, graph y = 150 as Y2 and determine where the two graphs intersect. Use the Intersect feature from the CALCULATE menu to determine where the two graphs intersect. Recall that interest is added at the end of each year; so, even though 4.254 163 7 is closer to 4 years, Anna must wait 5 years before her original investment will increase by at least 50%.

## Determining Doubling Time

Many people in the financial community can quickly estimate the time required for an investment earning compound interest to double in value by applying the “rule of 72.” They simply divide 72 by the annual interest rate. For example, an annual rate of 9% will double your money in about 72 ÷ 9 or 8 years. At 6%, your money will double in about 72 ÷ 6 or 12 years.

The “rule of 72” overestimates the time required for the principal to double if the interest rate is less than 8%. Likewise, it underestimates the time required for the principal to double if the interest rate is 8% or greater. The best estimates are when the interest rates are close to 8%.

 Example 1: Annual Calculation Money is invested at 6% per annum interest. How long will it take the investment to double if the interest is compounded anually. Solution Use the compound interest formula, A(t) = P(1 + i)n. In this question, i = 6% = 0.06. Because the interest is compounded annually, n = t. A(t) = P(1.06)t When the original amount doubles, A(t) = 2P. Because the interest is calculated annually, the time must be rounded up to 12 years. Therefore, it will take 12 years for the amount to at least double in value. Example 2: Monthly Calculation Money is invested at 6% per annum interest. How long will it take the investment to double if the interest is compounded monthly. Solution Use the compound interest formula, A(t) = P(1 + i)n. Because the interest is calculated monthly, the rate per period is Because the interest is compounded monthly, there are 12t, interest periods in t years. Therefore, n = 12t. A(t) = P(1.005)12t When the original amount doubles, A(t) = 2P. 0.6 years × 12 months/year = 7.2 months Because interest is calculated monthly, it will take 11 years 8 months for the original investment to double. Example 3: Semiannual Calculation Mark invests \$500 at a annual rate of 10% compounded semiannually. How long will it take the investment to double? Solution Method 1: A(t) = P(1 + i)n 1000 = 500(1 + 0.10/2)n 2 = (1 + 0.05)n log 2 = log 1.05n log 2 = n log 1.05 n =   log 2          log 1.05     14.2 Method 2: 1000 = 500(1 + 0.10/2)n Graph both sides of the equation on your graphing caluclator. Use the intersect feature to determine the value of n. n 14.2 Mark's investment will double in approximately 14 interest periods. Each period is half a year, so the time would be approximately 7 years.

## Calculating Present Value

 Example 1 Mark is putting money aside for a trip to Africa in 5 years. He estimates the trip will cost \$8000. If he can invest his money at 12% per year, compounded quarterly, how much must he invest now to reach his goal? Solution i = 12% = 0.03, A = \$8000, n = 5 · 4 = 20         4 A = P(1 + i)n Example 2   Solution

## Determining Depreciated Value

 Example 1 A \$3000 computer depreciates at a rate of 30% annually. What is the value of the computer after 4 years. Solution Because the annual depreciation is 30%, the value of the computer is 70% of the previous year’s value. Therefore, the value of the computer each year is as follows: Method 1 At the end of the first year, the computer is worth 70% of 3000 = 0.70 x 3000                      = \$2100 At the end of each successive year, the computer is worth 70% of its previous value. this is a geometric series with a = 2100, r = 0.070, n = 4 tn = arn-1    = 2100(0.70)4-1    = \$720.80 Method 2 After 4 years, Revenue Canada would assign a value of \$720.30 to the computer. Example 2 A car that initially costs \$32 000 depreciates at the rate of 15% each year. What is the value of the car after 3 years to the nearest dollar? Solution A = final value Ao = initial value i = rate of growth per period n = number of periods A = Ao(1 + i)n A = 32 000(1 + (-0.15))3     = 32 000(1 - 0.15)3     = 32 000(0.85)3     = \$19 652.00 The value of the car at the end of 3 years will be \$19 652.00.

Parts of this work has been adapted from a Math 30 Pure learning resource originally produced and owned by Alberta Education

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