**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

Answered: Image /qna-images/answer/19b806af-ba1a-4bd0-b917-3699a180d043.jpg

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 (*)✓✓✓.

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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Question

**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.

---

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