Question

When Nico was born his grandparents put $1000 in a special account t. The amount of money in the savings account increases by 8% every year.

Create a sequence that describes the amount of money in Nico’s savings each year for 8 years. (Write answers in the form or 2,000 or 1,567.87)

Answer 1

To create a sequence that describes the amount of money in Nico's savings account each year for 8 years, we start with the initial amount of $1000 and apply an annual increase of 8%.

The formula to calculate the amount of money in the account at the end of each year is given by:

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

where:
- \( A \) is the amount of money in the account after \( n \) years,
- \( P \) is the principal amount (initial amount),
- \( r \) is the annual interest rate (as a decimal),
- \( n \) is the number of years.

In this case:
- \( P = 1000 \)
- \( r = 0.08 \)

Now we can calculate the amount for each year from 1 to 8:

1. Year 1:
\[ A_1 = 1000(1 + 0.08)^1 = 1000 \times 1.08 = 1080.00 \]

2. Year 2:
\[ A_2 = 1000(1 + 0.08)^2 = 1000 \times 1.08^2 = 1000 \times 1.1664 = 1166.40 \]

3. Year 3:
\[ A_3 = 1000(1 + 0.08)^3 = 1000 \times 1.08^3 = 1000 \times 1.259712 = 1259.71 \]

4. Year 4:
\[ A_4 = 1000(1 + 0.08)^4 = 1000 \times 1.08^4 = 1000 \times 1.36049 = 1360.49 \]

5. Year 5:
\[ A_5 = 1000(1 + 0.08)^5 = 1000 \times 1.08^5 = 1000 \times 1.469328 = 1469.33 \]

6. Year 6:
\[ A_6 = 1000(1 + 0.08)^6 = 1000 \times 1.08^6 = 1000 \times 1.586874 = 1586.87 \]

7. Year 7:
\[ A_7 = 1000(1 + 0.08)^7 = 1000 \times 1.08^7 = 1000 \times 1.713778 = 1713.78 \]

8. Year 8:
\[ A_8 = 1000(1 + 0.08)^8 = 1000 \times 1.08^8 = 1000 \times 1.85093 = 1850.93 \]

Thus, the sequence that describes the amount of money in Nico's savings each year for 8 years is:

- Year 1: $1080.00
- Year 2: $1166.40
- Year 3: $1259.71
- Year 4: $1360.49
- Year 5: $1469.33
- Year 6: $1586.87
- Year 7: $1713.78
- Year 8: $1850.93

So, the sequence is:
**1080.00, 1166.40, 1259.71, 1360.49, 1469.33, 1586.87, 1713.78, 1850.93**