Suppose that you have $12,500 to invest over a 4 year period. There are two accounts to choose from: 4.5% compounded monthly or 4.3% compounded continuously.
a. Write the formula for the first account’s compound interest for n compounding per year.
b. Write the formula for the account with continuously compounded interest.
c. Use the formulas and information above to solve for the balance of each investment after 3 years.
d. Which investment account offers a better return on your money?
a. The formula for the first account's compound interest with n compounding per year is:
A = P (1 + r/n)^(nt)
Where:
A = the final account balance
P = the initial principal (in this case, $12,500)
r = the annual interest rate (in this case, 4.5%)
n = the number of times the interest is compounded per year
t = the number of years
b. The formula for the account with continuously compounded interest is:
A = P * e^(rt)
Where:
A = the final account balance
P = the initial principal (in this case, $12,500)
r = the annual interest rate (in this case, 4.3%)
t = the number of years
e = the mathematical constant approximately equal to 2.71828
c. To solve for the balance of each investment after 3 years, we can use the formulas from parts a and b:
For the first account with 4.5% compounded monthly:
A = 12,500 * (1 + 0.045/12)^(12*3)
Calculate the exponent first, then multiply it by the initial principal.
For the second account with 4.3% compounded continuously:
A = 12,500 * e^(0.043*3)
Multiply the annual interest rate by the number of years, then raise e (the base of natural logarithms) to that power, and finally multiply it by the initial principal.
d. Compare the balances obtained from the calculations in part c. The investment account that offers a better return on your money is the one with the higher final balance after 3 years.