Question

Write a geometric series using the following sequence that sums the first 100 terms in sigma notation:
4,−
3
4

,
9
4

,…

A.
n=1

100

(4)(−
3
1

)
n−1


B.
n=1

4

(100)(−
3
1

)
n−1


C.
n=1

100

(4)(−
9
1

)
n−1


D.
n=1

100

(−
3
1

)(4)
n−1

Answers

Answered by GPT-4o mini
To write the geometric series using sigma notation for the given sequence \( 4, -\frac{3}{4}, \frac{9}{4}, \ldots \), we first need to identify the first term (denote it as \( a \)) and the common ratio (denote it as \( r \)) of the series.

1. **Identify the first term**:
- The first term \( a \) is \( 4 \).

2. **Identify the common ratio**:
- To find the common ratio \( r \), divide the second term by the first term:
\[
r = \frac{-\frac{3}{4}}{4} = -\frac{3}{16}
\]

3. **Overall structure of a geometric series**:
- The sum of the first \( n \) terms of a geometric series can be expressed in sigma notation as:
\[
S_n = \sum_{n=1}^{k} ar^{n-1}
\]
where \( a \) is the first term and \( r \) is the common ratio.

From the discussion above, the series can be expressed as:
\[
\sum_{n=1}^{100} 4 \left(-\frac{3}{16}\right)^{n-1}
\]

Now, let's find the correct option among the choices given:

- **A:** \( \sum_{n=1}^{100} (4)(-\frac{3}{1})^{n-1} \) -> Incorrect common ratio.
- **B:** \( \sum_{n=1}^{100} 4(100)(-\frac{3}{1})^{n-1} \) -> Incorrect structure, wrong ratio and multiplier.
- **C:** \( \sum_{n=1}^{100} (4)(-\frac{9}{1})^{n-1} \) -> Also incorrect common ratio.
- **D:** \( \sum_{n=1}^{100} (-\frac{3}{1})(4)^{n-1} \) -> Also incorrect.

None of the options match the correct structure of the geometric series derived from the given sequence.

### Conclusion
None of the provided options correctly represent the geometric series for the sequence \( 4, -\frac{3}{4}, \frac{9}{4}, \ldots \). The correct sigma notation for this series with \( a = 4 \) and \( r = -\frac{3}{16} \) should be:
\[
\sum_{n=1}^{100} 4\left(-\frac{3}{16}\right)^{n-1}
\]

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