### Problem DescriptionFind the sum; write your answer as a fraction\[ \sum_{k=1}^{8} 3 \left(\frac{1}{4}\right)^{k-1} \]### ExplanationThis 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 Steps1. **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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Find the sum; write your answer as a fraction A-1 3. kw1

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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Question

### 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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