Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Problem Description
Find the sum; write your answer as a fraction
\[ \sum_{k=1}^{8} 3 \left(\frac{1}{4}\right)^{k-1} \]
### Explanation
This is a summation problem where you need to find the sum of a geometric series. The notation \(\sum_{k=1}^{8}\) indicates that we are summing from \(k=1\) to \(k=8\). Each term in the series is given by the formula \(3 \left(\frac{1}{4}\right)^{k-1}\).
### Solution Steps
1. **Identify the first term and common ratio:**
- First term (\(a\)): When \(k=1\), the term is \(3 \left(\frac{1}{4}\right)^{1-1} = 3(1) = 3\).
- Common ratio (\(r\)): The base of the exponent in the term formula is \(\frac{1}{4}\).
2. **Use the formula for the sum of first \(n\) terms of a geometric series:**
\[
S_n = \frac{a(1 - r^n)}{1 - r}
\]
Here, \(a = 3\), \(r = \frac{1}{4}\), and \(n = 8\).
3. **Substitute the values into the formula:**
\[
S_8 = \frac{3(1 - (\frac{1}{4})^8)}{1 - \frac{1}{4}}
\]
4. **Calculate the power and difference:**
\[
(\frac{1}{4})^8 = \frac{1}{65536}
\]
\[
1 - \frac{1}{65536} = \frac{65536 - 1}{65536} = \frac{65535}{65536}
\]
5. **Substitute back into the sum formula:**
\[
S_8 = \frac{3 \cdot \frac{65535}{65536}}{\frac{3}{4}} = \frac{3 \cdot 65535}{65536} \times \frac{4}{3}
\]
\[
S_8](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6c88b4b6-4152-4350-b43c-eadd21488152%2Fccbcf50d-0baa-403f-bff9-e4ecb539530b%2Fkvu8zsp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Description
Find the sum; write your answer as a fraction
\[ \sum_{k=1}^{8} 3 \left(\frac{1}{4}\right)^{k-1} \]
### Explanation
This is a summation problem where you need to find the sum of a geometric series. The notation \(\sum_{k=1}^{8}\) indicates that we are summing from \(k=1\) to \(k=8\). Each term in the series is given by the formula \(3 \left(\frac{1}{4}\right)^{k-1}\).
### Solution Steps
1. **Identify the first term and common ratio:**
- First term (\(a\)): When \(k=1\), the term is \(3 \left(\frac{1}{4}\right)^{1-1} = 3(1) = 3\).
- Common ratio (\(r\)): The base of the exponent in the term formula is \(\frac{1}{4}\).
2. **Use the formula for the sum of first \(n\) terms of a geometric series:**
\[
S_n = \frac{a(1 - r^n)}{1 - r}
\]
Here, \(a = 3\), \(r = \frac{1}{4}\), and \(n = 8\).
3. **Substitute the values into the formula:**
\[
S_8 = \frac{3(1 - (\frac{1}{4})^8)}{1 - \frac{1}{4}}
\]
4. **Calculate the power and difference:**
\[
(\frac{1}{4})^8 = \frac{1}{65536}
\]
\[
1 - \frac{1}{65536} = \frac{65536 - 1}{65536} = \frac{65535}{65536}
\]
5. **Substitute back into the sum formula:**
\[
S_8 = \frac{3 \cdot \frac{65535}{65536}}{\frac{3}{4}} = \frac{3 \cdot 65535}{65536} \times \frac{4}{3}
\]
\[
S_8
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