Math 30P Exponents Logarithms Sequences Series Regression/Recursive Interest Growth Richter/PH Prerequisite Skills
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
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}, …
Spreadsheet Step 2: Spreadsheet formulae:

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).
Step 7: Press to obtain the table.

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. SolutionCalculate 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.

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 1.05^{n} log 2 = n log 1.05 n = log 2 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. 

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 A = P(1 + i)^{n}

Example 2
Solution

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
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 = A_{o}(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
Math 30P Exponents Logarithms Sequences Series Regression/Recursive Interest Growth Richter/PH Prerequisite Skills