Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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Polynomials
Try substituting 1. You will get 6 and not 0. Try other real values also. What About -2? Let us check.
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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
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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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