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

Answer 1

To create a sequence that describes the amount of money in Nico’s savings account for each year, we can use the formula for exponential growth:

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

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

In this case:
- \( P = 1000 \) (the initial amount),
- \( r = 0.08 \) (8% interest rate),
- \( t \) will vary from 0 to 8 years.

Now we can calculate the amount for each year from 0 to 8:

1. Year 0:
\[
A_0 = 1000(1 + 0.08)^0 = 1000(1) = 1000
\]

2. Year 1:
\[
A_1 = 1000(1 + 0.08)^1 = 1000(1.08) = 1080
\]

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

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

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

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

7. Year 6:
\[
A_6 = 1000(1 + 0.08)^6 = 1000(1.08)^6 \approx 1587.14
\]

8. Year 7:
\[
A_7 = 1000(1 + 0.08)^7 = 1000(1.08)^7 \approx 1714.58
\]

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

Now, we can summarize the amounts for each year in a sequence:

**Sequence of savings amounts for the first 8 years:**
- Year 0: $1000
- Year 1: $1080
- Year 2: $1166.40
- Year 3: $1259.71
- Year 4: $1360.49
- Year 5: $1469.33
- Year 6: $1587.14
- Year 7: $1714.58
- Year 8: $1850.93

Thus, the sequence (rounded to two decimal places) is:
\[
1000, 1080, 1166.40, 1259.71, 1360.49, 1469.33, 1587.14, 1714.58, 1850.93
\]