To describe the amount of money in Nico's savings account over time, we can use the formula for compound interest:
\[ A = P (1 + r)^t \]
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).
- \( t \) is the number of years the money is invested or borrowed for.
In this case:
- \( P = 1000 \)
- \( r = 0.08 \) (which is 8%)
- \( t \) will vary from 0 to 8.
We'll calculate the amount for each year from 0 to 8:
-
Year 0 (initial deposit): \[ A = 1000 \times (1 + 0.08)^0 = 1000 \times 1 = 1000 \]
-
Year 1: \[ A = 1000 \times (1 + 0.08)^1 = 1000 \times 1.08 = 1080 \]
-
Year 2: \[ A = 1000 \times (1 + 0.08)^2 = 1000 \times 1.1664 = 1166.40 \]
-
Year 3: \[ A = 1000 \times (1 + 0.08)^3 = 1000 \times 1.259712 = 1259.71 \]
-
Year 4: \[ A = 1000 \times (1 + 0.08)^4 = 1000 \times 1.36048864 = 1360.49 \]
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Year 5: \[ A = 1000 \times (1 + 0.08)^5 = 1000 \times 1.469328 = 1469.33 \]
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Year 6: \[ A = 1000 \times (1 + 0.08)^6 = 1000 \times 1.586874 = 1586.87 \]
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Year 7: \[ A = 1000 \times (1 + 0.08)^7 = 1000 \times 1.713775 = 1713.78 \]
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Year 8: \[ A = 1000 \times (1 + 0.08)^8 = 1000 \times 1.85093 = 1850.93 \]
Now, compiling this into a sequence for each year from 0 to 8:
- Year 0: 1000.00
- 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
Thus, the sequence of the amount in Nico’s savings account each year for 8 years is:
1000.00, 1080.00, 1166.40, 1259.71, 1360.49, 1469.33, 1586.87, 1713.78, 1850.93