Question
A(t)= p(1+ r/n)^nt Marion deposits $100 into a savings account that has a 6% annualized interest rate. The bank offers four compounding
schedules: annually, quarterly, weekly, and daily. Show which plan provides Marion with the best offer. Assume the
money was deposited for 1 year. How much more money does Marion accrue if it is compounded daily rather than
annually?
schedules: annually, quarterly, weekly, and daily. Show which plan provides Marion with the best offer. Assume the
money was deposited for 1 year. How much more money does Marion accrue if it is compounded daily rather than
annually?
Answers
GPT-4o mini
To determine which compounding schedule provides Marion with the best offer, we'll use the compound interest formula:
\[
A(t) = P \left( 1 + \frac{r}{n} \right)^{nt}
\]
Where:
- \( A(t) \) is the amount of money accumulated after n years, including interest.
- \( P \) is the principal amount (the initial amount of money, which in this case is $100).
- \( r \) is the annual interest rate (decimal).
- \( n \) is the number of times that interest is compounded per year.
- \( t \) is the number of years the money is invested or borrowed.
In this scenario:
- \( P = 100 \)
- \( r = 0.06 \) (6% annual interest)
- \( t = 1 \) year
Now, we'll calculate \( A(t) \) for each of the compounding frequencies:
### 1. Compounded Annually (\( n = 1 \)):
\[
A(1) = 100 \left( 1 + \frac{0.06}{1} \right)^{1 \cdot 1} = 100 \left( 1 + 0.06 \right)^1 = 100 \times 1.06 = 106
\]
### 2. Compounded Quarterly (\( n = 4 \)):
\[
A(1) = 100 \left( 1 + \frac{0.06}{4} \right)^{4 \cdot 1} = 100 \left( 1 + 0.015 \right)^4 = 100 \left( 1.015 \right)^4
\]
Calculating \( (1.015)^4 \):
\[
(1.015)^4 \approx 1.061364
\]
\[
A(1) \approx 100 \times 1.061364 = 106.1364
\]
### 3. Compounded Weekly (\( n = 52 \)):
\[
A(1) = 100 \left( 1 + \frac{0.06}{52} \right)^{52 \cdot 1} = 100 \left( 1 + \frac{0.06}{52} \right)^{52} = 100 \left( 1 + 0.001153846 \right)^{52}
\]
Calculating \( (1.001153846)^{52} \):
\[
(1.001153846)^{52} \approx 1.061677812
\]
\[
A(1) \approx 100 \times 1.061677812 \approx 106.16778
\]
### 4. Compounded Daily (\( n = 365 \)):
\[
A(1) = 100 \left( 1 + \frac{0.06}{365} \right)^{365 \cdot 1} = 100 \left( 1 + \frac{0.06}{365} \right)^{365}
\]
Calculating \( (1 + \frac{0.06}{365})^{365} \):
\[
\left( 1 + \frac{0.06}{365} \right)^{365} \approx 1.061736
\]
\[
A(1) \approx 100 \times 1.061736 = 106.1736
\]
### Summary of Results:
- Compounded Annually: \( A = 106.00 \)
- Compounded Quarterly: \( A \approx 106.1364 \)
- Compounded Weekly: \( A \approx 106.16778 \)
- Compounded Daily: \( A \approx 106.1736 \)
### Difference Between Daily and Annually:
To find how much more money Marion accrues by compounding daily compared to annually:
\[
\text{Difference} = A_{\text{Daily}} - A_{\text{Annually}} = 106.1736 - 106 = 0.1736
\]
### Final Answer:
Marion accrues approximately **$0.17** more with daily compounding compared to annual compounding after one year. The best compounding schedule is daily compounding.
\[
A(t) = P \left( 1 + \frac{r}{n} \right)^{nt}
\]
Where:
- \( A(t) \) is the amount of money accumulated after n years, including interest.
- \( P \) is the principal amount (the initial amount of money, which in this case is $100).
- \( r \) is the annual interest rate (decimal).
- \( n \) is the number of times that interest is compounded per year.
- \( t \) is the number of years the money is invested or borrowed.
In this scenario:
- \( P = 100 \)
- \( r = 0.06 \) (6% annual interest)
- \( t = 1 \) year
Now, we'll calculate \( A(t) \) for each of the compounding frequencies:
### 1. Compounded Annually (\( n = 1 \)):
\[
A(1) = 100 \left( 1 + \frac{0.06}{1} \right)^{1 \cdot 1} = 100 \left( 1 + 0.06 \right)^1 = 100 \times 1.06 = 106
\]
### 2. Compounded Quarterly (\( n = 4 \)):
\[
A(1) = 100 \left( 1 + \frac{0.06}{4} \right)^{4 \cdot 1} = 100 \left( 1 + 0.015 \right)^4 = 100 \left( 1.015 \right)^4
\]
Calculating \( (1.015)^4 \):
\[
(1.015)^4 \approx 1.061364
\]
\[
A(1) \approx 100 \times 1.061364 = 106.1364
\]
### 3. Compounded Weekly (\( n = 52 \)):
\[
A(1) = 100 \left( 1 + \frac{0.06}{52} \right)^{52 \cdot 1} = 100 \left( 1 + \frac{0.06}{52} \right)^{52} = 100 \left( 1 + 0.001153846 \right)^{52}
\]
Calculating \( (1.001153846)^{52} \):
\[
(1.001153846)^{52} \approx 1.061677812
\]
\[
A(1) \approx 100 \times 1.061677812 \approx 106.16778
\]
### 4. Compounded Daily (\( n = 365 \)):
\[
A(1) = 100 \left( 1 + \frac{0.06}{365} \right)^{365 \cdot 1} = 100 \left( 1 + \frac{0.06}{365} \right)^{365}
\]
Calculating \( (1 + \frac{0.06}{365})^{365} \):
\[
\left( 1 + \frac{0.06}{365} \right)^{365} \approx 1.061736
\]
\[
A(1) \approx 100 \times 1.061736 = 106.1736
\]
### Summary of Results:
- Compounded Annually: \( A = 106.00 \)
- Compounded Quarterly: \( A \approx 106.1364 \)
- Compounded Weekly: \( A \approx 106.16778 \)
- Compounded Daily: \( A \approx 106.1736 \)
### Difference Between Daily and Annually:
To find how much more money Marion accrues by compounding daily compared to annually:
\[
\text{Difference} = A_{\text{Daily}} - A_{\text{Annually}} = 106.1736 - 106 = 0.1736
\]
### Final Answer:
Marion accrues approximately **$0.17** more with daily compounding compared to annual compounding after one year. The best compounding schedule is daily compounding.