Use index substitution to rewrite the following summation such that the index starts at 0. Then use the geometric series theorem to compute the value of the summation. 6 Σ (2) 4 i=-2
Use index substitution to rewrite the following summation such that the index starts at 0. Then use the geometric series theorem to compute the value of the summation. 6 Σ (2) 4 i=-2
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
![**Instruction on Rewriting Summations Using Index Substitution**
**Objective:**
Learn how to rewrite a given summation so that the index starts at 0, and use the geometric series theorem to compute the value of the summation.
**Problem Statement:**
The original summation is expressed as:
\[
\sum_{i=-2}^{6} (2)^{i-4}
\]
**Steps to Solve:**
1. **Index Substitution:**
- The goal is to rewrite the sum such that the index starts at 0.
- Let \( j = i + 2 \). Then when \( i = -2 \), \( j = 0 \); and when \( i = 6 \), \( j = 8 \).
2. **Update the Summation:**
- Substitute \( i \) with \( j - 2 \) in the original expression.
- The summation now becomes:
\[
\sum_{j=0}^{8} (2)^{(j-2)-4} = \sum_{j=0}^{8} (2)^{j-6}
\]
3. **Geometric Series Theorem:**
- Recognize this as a geometric series of the form:
\[
a + ar + ar^2 + \cdots + ar^n
\]
- Here, \( a = (2)^{-6} \) and \( r = 2 \).
- Use the formula for the sum of a geometric series:
\[
S_n = a \frac{r^{n+1} - 1}{r - 1}
\]
- Calculate using \( a = (2)^{-6} \), \( r = 2 \), and \( n = 8 \).
4. **Compute the Sum:**
- Input these values into the formula to find the numerical value for the given summation.
By following these steps, you will be able to correctly rewrite and solve the summation using elementary transformations and the geometric series theorem.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2e521a1e-f4ee-4c02-af71-8dfc2cc184d3%2F7f651d76-66a0-45b1-b4e9-64bc48335196%2F45juebf_processed.png&w=3840&q=75)
Transcribed Image Text:**Instruction on Rewriting Summations Using Index Substitution**
**Objective:**
Learn how to rewrite a given summation so that the index starts at 0, and use the geometric series theorem to compute the value of the summation.
**Problem Statement:**
The original summation is expressed as:
\[
\sum_{i=-2}^{6} (2)^{i-4}
\]
**Steps to Solve:**
1. **Index Substitution:**
- The goal is to rewrite the sum such that the index starts at 0.
- Let \( j = i + 2 \). Then when \( i = -2 \), \( j = 0 \); and when \( i = 6 \), \( j = 8 \).
2. **Update the Summation:**
- Substitute \( i \) with \( j - 2 \) in the original expression.
- The summation now becomes:
\[
\sum_{j=0}^{8} (2)^{(j-2)-4} = \sum_{j=0}^{8} (2)^{j-6}
\]
3. **Geometric Series Theorem:**
- Recognize this as a geometric series of the form:
\[
a + ar + ar^2 + \cdots + ar^n
\]
- Here, \( a = (2)^{-6} \) and \( r = 2 \).
- Use the formula for the sum of a geometric series:
\[
S_n = a \frac{r^{n+1} - 1}{r - 1}
\]
- Calculate using \( a = (2)^{-6} \), \( r = 2 \), and \( n = 8 \).
4. **Compute the Sum:**
- Input these values into the formula to find the numerical value for the given summation.
By following these steps, you will be able to correctly rewrite and solve the summation using elementary transformations and the geometric series theorem.
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