It follows that for every integer k ≥ 1, 4bk- k-1 = Thus, bo, b₁,b₂, Need Help? 4k-1-3 4k x Read It x X ) by substitution from (*)✔✔ by basic algebra ... satisfies the given recurrence relation. D by substitution from (*)✓✓✓.

Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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**Transcription for Educational Website**

---

**Problem Description:**

Let \( b_0, b_1, b_2, \ldots \) be defined by the formula \( b_n = 4^n \) for every integer \( n \geq 0 \). Fill in the blanks to show that \( b_0, b_1, b_2, \ldots \) satisfies the recurrence relation \( b_k = 4b_{k-1} - 1 \) for every integer \( k \geq 1 \).

**Solution Steps:**

1. **Define \( b_k \) and \( b_{k-1} \):**

   - Let \( k \) be any integer with \( k \geq 1 \).
   - Substitute \( k \) and \( k-1 \) in place of \( n \), applying the definition of \( b_0, b_1, b_2, \ldots \) to both \( b_k \) and \( b_{k-1} \).
   - The results are:

     \[
     b_k = 4^k \quad (*) \quad \text{and} \quad b_{k-1} = 4
     \]
     (** refers to these definitions for every \( k \geq 1 \).

2. **Derive the Recurrence Relation:**

   - It follows that for every integer \( k \geq 1 \),

     \[
     4b_{k-1} - 1 = 4 \left( k - 1 \right) \quad \text{(by substitution from \((*)\))}
     \]

   - Simplifying using basic algebra gives:

     \[
     4(k - 1) - 3 
     \]

   - Finally, substitute from the earlier definition:

     \[
     4k = b_k
     \]

3. **Conclusion:**

   Thus, \( b_0, b_1, b_2, \ldots \) satisfies the given recurrence relation.

---

**Additional Resources:**
- [Need Help?](#) - Click here for further explanations.

**Submission:**
- [Submit Answer](#)

**Navigation:**
- [Details](#)
- [Previous Answers](#)
- My Notes: [My Notes](#)
- **Ask
Transcribed Image Text:**Transcription for Educational Website** --- **Problem Description:** Let \( b_0, b_1, b_2, \ldots \) be defined by the formula \( b_n = 4^n \) for every integer \( n \geq 0 \). Fill in the blanks to show that \( b_0, b_1, b_2, \ldots \) satisfies the recurrence relation \( b_k = 4b_{k-1} - 1 \) for every integer \( k \geq 1 \). **Solution Steps:** 1. **Define \( b_k \) and \( b_{k-1} \):** - Let \( k \) be any integer with \( k \geq 1 \). - Substitute \( k \) and \( k-1 \) in place of \( n \), applying the definition of \( b_0, b_1, b_2, \ldots \) to both \( b_k \) and \( b_{k-1} \). - The results are: \[ b_k = 4^k \quad (*) \quad \text{and} \quad b_{k-1} = 4 \] (** refers to these definitions for every \( k \geq 1 \). 2. **Derive the Recurrence Relation:** - It follows that for every integer \( k \geq 1 \), \[ 4b_{k-1} - 1 = 4 \left( k - 1 \right) \quad \text{(by substitution from \((*)\))} \] - Simplifying using basic algebra gives: \[ 4(k - 1) - 3 \] - Finally, substitute from the earlier definition: \[ 4k = b_k \] 3. **Conclusion:** Thus, \( b_0, b_1, b_2, \ldots \) satisfies the given recurrence relation. --- **Additional Resources:** - [Need Help?](#) - Click here for further explanations. **Submission:** - [Submit Answer](#) **Navigation:** - [Details](#) - [Previous Answers](#) - My Notes: [My Notes](#) - **Ask
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