Question

Question
A computer programmer makes $60,000 in her first year of working at a company. She gets a 10 percent pay raise every year. Create a geometric series model for how much she makes in his first 5 years of working.(1 point)
Responses

∑5n=160,000(1.1)n−1
sigma-summation underscript n equals 1 overscript 5 End Scripts 60,000 left parenthesis 1.1 right parenthesis superscript n minus 1 baseline

∑5n=160,000(0.1)n−1
sigma-summation underscript n equals 1 overscript 5 End Scripts 60,000 left parenthesis 0.1 right parenthesis superscript n minus 1 baseline

∑5n=560,000(1.1)n
sigma-summation underscript n equals 5 overscript 5 End Scripts 60,000 left parenthesis 1.1 right parenthesis superscript n baseline

∑5n=360,000(1.1)n−1

Answers

GPT-4o mini
To create a geometric series model for how much a computer programmer makes in her first 5 years of working, we start with her initial salary and the fact that she receives a 10% pay raise each year.

1. **Year 1:** Starts at $60,000.
2. **Year 2:** $60,000 * (1 + 0.10) = $60,000 * 1.1 = $66,000.
3. **Year 3:** $66,000 * 1.1 = $60,000 * (1.1)^2.
4. **Year 4:** $60,000 * (1.1)^3.
5. **Year 5:** $60,000 * (1.1)^4.

The total amount earned in the first 5 years can be represented as a geometric series:

\[
\text{Total Earnings} = 60,000 + 60,000 \cdot 1.1 + 60,000 \cdot (1.1)^2 + 60,000 \cdot (1.1)^3 + 60,000 \cdot (1.1)^4
\]

This can be factored to represent a summation:

\[
\text{Total Earnings} = 60,000 \sum_{n=0}^{4} (1.1)^n
\]

To convert the summation index to start from 1 (as is typical in your choices), it can be rewritten as:

\[
\text{Total Earnings} = 60,000 \sum_{n=1}^{5} (1.1)^{n-1}
\]

Now let’s check which option matches this:

1. \(\sum_{n=1}^{5} 60,000(1.1)^{n-1}\) - This matches our model.
2. \(\sum_{n=1}^{5} 60,000(0.1)^{n-1}\) - This is incorrect.
3. \(\sum_{n=5}^{5} 60,000(1.1)^{n}\) - This doesn't represent 5 terms correctly.
4. \(\sum_{n=1}^{5} 60,000(1.1)^{n-1}\) - This matches our model.

Thus, the correct answer is:

\[
\sum_{n=1}^{5} 60,000(1.1)^{n-1}
\]

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