Questions: Suppose that a loan of 3000 is given at an interest rate of 6% compounded each year. Assume that no payments are made on the loan. (a) Find the amount owed at the end of 1 year. (b) Find the amount owed at the end of 2 years.

Solution Steps

To solve this problem, we need to use the formula for compound interest, which is given by:

\[ A = P(1 + r)^n \]

where:

• \( A \) is the amount of money accumulated after n years, including interest.
• \( P \) is the principal amount (the initial amount of money).
• \( r \) is the annual interest rate (decimal).
• \( n \) is the number of years the money is invested or borrowed for.

(a) For the amount owed at the end of 1 year, we substitute \( P = 3000 \), \( r = 0.06 \), and \( n = 1 \) into the formula.

(b) For the amount owed at the end of 2 years, we substitute \( P = 3000 \), \( r = 0.06 \), and \( n = 2 \) into the formula.

To find the amount owed on a loan with compound interest, we use the formula:

\[ A = P(1 + r)^n \]

where:

• \( A \) is the amount after \( n \) years,
• \( P \) is the principal amount,
• \( r \) is the annual interest rate (as a decimal),
• \( n \) is the number of years.

For the first year, we substitute the given values into the formula:

• \( P = 3000 \)
• \( r = 0.06 \)
• \( n = 1 \)

\[ A_1 = 3000(1 + 0.06)^1 = 3000 \times 1.06 = 3180.0 \]

For the second year, we use the same formula with \( n = 2 \):

\[ A_2 = 3000(1 + 0.06)^2 = 3000 \times 1.1236 = 3370.8 \]

Final Answer