Question
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rite the geometric series \( 9+3+1 ; \ldots \) to 130 terms in sigma notati

Ask by Malone Blake. in South Africa
Feb 03,2025

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Answer

\[ \sum_{n=1}^{130} 9 \cdot \left(\frac{1}{3}\right)^{n-1} \]

Solution

To express the geometric series \( 9 + 3 + 1 + \ldots \) up to 130 terms using sigma notation, follow these steps: 1. **Identify the first term and the common ratio:** - **First term (\( a \))**: 9 - **Common ratio (\( r \))**: \( \frac{3}{9} = \frac{1}{3} \) 2. **Determine the general term of the series:** The \( n \)-th term of a geometric series can be expressed as: \[ a \cdot r^{n-1} \] For this series, it becomes: \[ 9 \cdot \left( \frac{1}{3} \right)^{n-1} \] 3. **Write the sigma notation:** Summing up the first 130 terms, the series can be written as: \[ \sum_{n=1}^{130} 9 \cdot \left( \frac{1}{3} \right)^{n-1} \] **Final Answer:** \[ \sum_{n=1}^{130} 9 \cdot \left(\frac{1}{3}\right)^{\,n-1} \]

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The Deep Dive

The geometric series \( 9 + 3 + 1 + \ldots \) can be expressed in sigma notation as follows: \[ \sum_{n=0}^{129} 9 \left( \frac{1}{3} \right)^n \] Here, the first term \( a = 9 \) and the common ratio \( r = \frac{1}{3} \). The upper limit of 129 accounts for a total of 130 terms.

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