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 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:
\[
S_{15} = \frac{15}{2} \times (-8)
\]
\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F159ba596-6a8b-4dd1-b047-54789f82ea1f%2F76bd7ddb-b7da-4ede-b6a5-1e016e308830%2Fjhr1ntb.jpeg&w=3840&q=75)
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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