Problem 71 - Precalculus Mathematics for Calculus

Question

Julio deposits \(\$ 2000\) in a savings account that pays \(2.4 \%\) interest per year compounded monthly. The amount in the account after \(n\) months is given by the sequence $$ A_{n}=2000\left(1+\frac{0.024}{12}\right)^{n} $$ (a) Find the first six terms of the sequence. (b) Find the amount in the account after 3 years.

Step-by-Step Solution

Answer

(a) Terms: \$2004, \$2008.01, \$2012.02, \$2016.05, \$2020.08, \$2024.12. (b) \$2148.85 after 3 years.

Step 1: Understand the Formula

The given formula for the amount in Julio's savings account after \(n\) months is \[ A_{n} = 2000 \left(1 + \frac{0.024}{12}\right)^{n} \] which includes an initial deposit of \( \$2000 \) and a monthly compounding interest rate. The monthly interest rate is \( \frac{0.024}{12} = 0.002 \).

Step 2: Compute the First Term (n=1)

To find the first term of the sequence, substitute \( n = 1 \) into the formula: \[ A_1 = 2000 \left(1 + 0.002\right)^1 = 2000 \times 1.002 = 2004 \. \] Thus, the first term is \( \$2004 \).

Step 3: Compute the Second Term (n=2)

Substitute \( n = 2 \) into the formula to find the second term: \[ A_2 = 2000 \left(1.002\right)^2 = 2000 \times 1.004004 = 2008.008 \approx 2008.01 \] Thus, the second term is approximately \( \$2008.01 \).

Step 4: Compute the Third Term (n=3)

Substitute \( n = 3 \) into the formula to find the third term: \[ A_3 = 2000 \left(1.002\right)^3 = 2000 \times 1.006012008 = 2012.024016 \approx 2012.02 \] Thus, the third term is approximately \( \$2012.02 \).

Step 5: Compute the Fourth Term (n=4)

Substitute \( n = 4 \) into the formula to find the fourth term: \[ A_4 = 2000 \left(1.002\right)^4 = 2000 \times 1.008024 = 2016.048 \approx 2016.05 \] Thus, the fourth term is approximately \( \$2016.05 \).

Step 6: Compute the Fifth Term (n=5)

Substitute \( n = 5 \) into the formula to find the fifth term: \[ A_5 = 2000 \left(1.002\right)^5 = 2000 \times 1.010040048 = 2020.080096 \approx 2020.08 \] Thus, the fifth term is approximately \( \$2020.08 \).

Step 7: Compute the Sixth Term (n=6)

Substitute \( n = 6 \) into the formula to find the sixth term: \[ A_6 = 2000 \left(1.002\right)^6 = 2000 \times 1.012060128 = 2024.12024 \approx 2024.12 \] Thus, the sixth term is approximately \( \$2024.12 \).

Step 8: Convert Years to Months (3 Years)

To find the amount after 3 years, we need to convert the years to months: \( 3 \text{ years} = 3 \times 12 = 36 \text{ months} \).

Step 9: Compute the Amount After 3 Years (n=36)

Substitute \( n = 36 \) into the formula to find the total amount after 3 years: \[ A_{36} = 2000 \left(1.002\right)^{36} \approx 2000 \times 1.074424674 \approx 2148.85 \] Thus, the amount in the account after 3 years is approximately \( \$2148.85 \).

Key Concepts

Sequences

Sequences are ordered lists of numbers where each term follows a specific rule or pattern. In the context of compound interest, we deal with sequences that represent the growth of an investment over time. For Julio's savings account, the sequence is defined by the formula \[ A_{n} = 2000 \left(1 + \frac{0.024}{12}\right)^{n} \] Each term of this sequence represents the amount of money in the account after a specific number of months. Sequentially, for each subsequent month's term, we calculate the accumulated amount by raising the monthly interest rate to the power of the number of months, making this a geometric sequence because of the constant ratio between consecutive terms.

Monthly Compounding

Monthly compounding refers to the process of calculating interest on an investment and adding that interest to the principal every month. This means that each month not only does the principal earn interest, but so does the previously accumulated interest.

• The annual interest rate in Julio's account is 2.4%.
• For monthly compounding, we divide this rate by 12 to get the monthly rate:

\[ \text{Monthly Interest Rate} = \frac{0.024}{12} = 0.002 \] Over time, this results in exponential growth as more interest is calculated on top of past interest amounts. Monthly compounding can lead to a significantly increased amount compared to annual compounding, because interest starts being calculated earlier on the reinvested gains.

Interest Calculations

The formula for calculating interest accumulations in a savings account involves several steps. Let's break it down:

• Start with an initial principal amount. In our case, it's \$2000.
• Use the formula: \[ A_{n} = P \left(1 + r\right)^{n} \]
• Here, \( P \) is the principal, \( r \) is the monthly interest rate (0.002), and \( n \) is the number of months.
• For each term \( A_n \), substitute \( n \) to get that month's total amount.

For instance, the first term when \( n = 1 \) becomes: \[ A_1 = 2000 \times 1.002 \approx 2004 \] The repetitive nature of the calculation illustrates how the amount compounds monthly.

Savings Account Growth

Savings account growth in Julio's scenario is an example of how small, regular increases in a principal can accumulate over time due to compound interest. The principal grows more significantly than simple interest because each month's earnings are added to the principal for the next month's interest calculation. After three years, or 36 months, the effect of compounding becomes more pronounced, and the total amount grows considerably: \[ A_{36} \approx 2148.85 \] This shows that compound interest is a powerful tool for growing savings because the compounded earnings themselves generate further interest, continuously increasing the savings account balance.