Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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Set Theory
Every field in the present world involves a set. The theory of sets was developed by German mathematician Georg Cantor while working on the “problems of trigonometric series”.
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### Problem Statement **Part 2. Find the sum:** \[ \sum_{k=1}^{15} (20 - 3k) \] ### Explanation This mathematical notation represents the sum of the series defined by the expression \( 20 - 3k \) as \( k \) runs from 1 to 15. To compute this, we need to substitute each integer value of \( k \) from 1 to 15 into the expression \( 20 - 3k \) and then add up all those values. Here's how you can break it down step-by-step: 1. Substitute \( k = 1 \) into \( 20 - 3k \). 2. Substitute \( k = 2 \) into \( 20 - 3k \). 3. Continue this process up to \( k = 15 \). 4. Sum up all the resulting values: \[ (20 - 3 \cdot 1) + (20 - 3 \cdot 2) + (20 - 3 \cdot 3) + \ldots + (20 - 3 \cdot 15) \] ### Series Summation We notice that this is an arithmetic series where the first term \( a = 20 - 3 \times 1 = 17 \) and the common difference \( d = -3 \). The number of terms \( n = 15 \). The sum \( S_n \) of an arithmetic series can be calculated using the formula: \[ S_n = \frac{n}{2} \times (2a + (n-1)d) \] Substituting the values: \[a = 17,\ d = -3,\ \text{and}\ n = 15:\] \[ S_{15} = \frac{15}{2} \times [2 \times 17 + (15 - 1) \times (-3)] \] Solving inside the brackets first: \[ 2 \times 17 = 34 \] \[ (15 - 1) \times (-3) = 14 \times (-3) = -42 \] Then: \[ 34 + (-42) = -8 \] Now, apply the summation formula: \[ S_{15} = \frac{15}{2} \times (-8) \] \
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Transcribed Image Text:### Problem Statement **Part 2. Find the sum:** \[ \sum_{k=1}^{15} (20 - 3k) \] ### Explanation This mathematical notation represents the sum of the series defined by the expression \( 20 - 3k \) as \( k \) runs from 1 to 15. To compute this, we need to substitute each integer value of \( k \) from 1 to 15 into the expression \( 20 - 3k \) and then add up all those values. Here's how you can break it down step-by-step: 1. Substitute \( k = 1 \) into \( 20 - 3k \). 2. Substitute \( k = 2 \) into \( 20 - 3k \). 3. Continue this process up to \( k = 15 \). 4. Sum up all the resulting values: \[ (20 - 3 \cdot 1) + (20 - 3 \cdot 2) + (20 - 3 \cdot 3) + \ldots + (20 - 3 \cdot 15) \] ### Series Summation We notice that this is an arithmetic series where the first term \( a = 20 - 3 \times 1 = 17 \) and the common difference \( d = -3 \). The number of terms \( n = 15 \). The sum \( S_n \) of an arithmetic series can be calculated using the formula: \[ S_n = \frac{n}{2} \times (2a + (n-1)d) \] Substituting the values: \[a = 17,\ d = -3,\ \text{and}\ n = 15:\] \[ S_{15} = \frac{15}{2} \times [2 \times 17 + (15 - 1) \times (-3)] \] Solving inside the brackets first: \[ 2 \times 17 = 34 \] \[ (15 - 1) \times (-3) = 14 \times (-3) = -42 \] Then: \[ 34 + (-42) = -8 \] Now, apply the summation formula: \[ S_{15} = \frac{15}{2} \times (-8) \] \
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