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
How do I solve #3?
![**Summation Formulas and Evaluations**
The image presents summation properties and formulas to evaluate specific finite series. Here's a detailed transcription of the content:
### Summation Formulas:
1. \( \sum_{i=1}^{n} i = \frac{n(n+1)}{2} \)
2. \( \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \)
3. \( \sum_{i=1}^{n} i^3 = \left(\frac{n(n+1)}{2}\right)^2 \)
4. \( \sum_{i=0}^{n} r^i = \frac{1-r^{n+1}}{1-r}, \; r \neq 1 \) (Geometric Sum)
5. \( \sum_{i=1}^{n} 1 = n \)
### Problem 6: Summation Evaluation
**Use summation properties and formula to rewrite and evaluate the sums for each of the following finite series:**
1. **\[ \sum_{k=-2}^{20} 100(k^2 - 5k + 1) \]**
To solve, rewrite and evaluate by adjusting indices:
- \( k + 2 = i \)
- Increment k from -2 to 20 translates into incrementing i from 1 to 23.
The expression becomes:
\[
100 \left(\sum_{i=1}^{23} (i^2 - 3i + 1)\right)
\]
Simplifies to:
\[
100 \left(\sum_{i=1}^{23} i^2 - 11i + 25 \right)
\]
Using summation formulas:
\[
100 \left( \frac{23(23+1)(2 \times 23+1)}{6} - 11 \times \frac{23(23+1)}{2} + 25 \times 23\right)
\]
Final result is 186300.
2. **\[ \sum_{k=12}^{20} (k^2 - 2k) \]**
Not evaluated in the transcription.
3. **\[ \frac{\sum_{j=5}^{15} (](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb234898a-ca00-4740-ae3d-79576f672843%2Ff3a30abb-6b1c-4ac7-966d-adf5b63a4eb7%2Fzk63dwm_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Summation Formulas and Evaluations**
The image presents summation properties and formulas to evaluate specific finite series. Here's a detailed transcription of the content:
### Summation Formulas:
1. \( \sum_{i=1}^{n} i = \frac{n(n+1)}{2} \)
2. \( \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \)
3. \( \sum_{i=1}^{n} i^3 = \left(\frac{n(n+1)}{2}\right)^2 \)
4. \( \sum_{i=0}^{n} r^i = \frac{1-r^{n+1}}{1-r}, \; r \neq 1 \) (Geometric Sum)
5. \( \sum_{i=1}^{n} 1 = n \)
### Problem 6: Summation Evaluation
**Use summation properties and formula to rewrite and evaluate the sums for each of the following finite series:**
1. **\[ \sum_{k=-2}^{20} 100(k^2 - 5k + 1) \]**
To solve, rewrite and evaluate by adjusting indices:
- \( k + 2 = i \)
- Increment k from -2 to 20 translates into incrementing i from 1 to 23.
The expression becomes:
\[
100 \left(\sum_{i=1}^{23} (i^2 - 3i + 1)\right)
\]
Simplifies to:
\[
100 \left(\sum_{i=1}^{23} i^2 - 11i + 25 \right)
\]
Using summation formulas:
\[
100 \left( \frac{23(23+1)(2 \times 23+1)}{6} - 11 \times \frac{23(23+1)}{2} + 25 \times 23\right)
\]
Final result is 186300.
2. **\[ \sum_{k=12}^{20} (k^2 - 2k) \]**
Not evaluated in the transcription.
3. **\[ \frac{\sum_{j=5}^{15} (
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