(5) If the series {a} satisfies that a₁ = 1.0₂ = 2, and a, – 40 1 + [1-5] (k > 1). [1-6] then ak =
(5) If the series {a} satisfies that a₁ = 1.0₂ = 2, and a, – 40 1 + [1-5] (k > 1). [1-6] then ak =
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section: Chapter Questions
Problem 22RE
Related questions
Question
![### Problem Statement
**(5)** If the series \(\{a_k\}\) satisfies that \(a_1 = 1\), \(a_2 = 2\), and \(a_k - 4a_{k-1} + 3a_{k-2} = 0\) (\(k \geq 3\)), then:
\[ a_k = \frac{1 + \left\lfloor \frac{k-5}{k-6} \right\rfloor}{\left\lfloor \frac{1-5}{1-6} \right\rfloor} \quad (k \geq 1) \]
### Explanation
This problem is related to solving a recurrence relation for the series \(\{a_k\}\).
- The initial conditions are given as \(a_1 = 1\) and \(a_2 = 2\).
- The recurrence relation is \(a_k - 4a_{k-1} + 3a_{k-2} = 0\) for \(k \geq 3\).
The provided formula \(a_k = \frac{1 + \left\lfloor \frac{k-5}{k-6} \right\rfloor}{\left\lfloor \frac{1-5}{1-6} \right\rfloor} \quad (k \geq 1)\) seems to involve floor functions. However, if intended to represent a general solution to the recurrence relation, the floor functions may simplify further.
### Detailed Steps:
1. **Initialize the series:**
- \( a_1 = 1 \)
- \( a_2 = 2 \)
2. **Recurrence relation:**
\[
a_k = 4a_{k-1} - 3a_{k-2} \quad \text{for} \quad k \geq 3
\]
3. **Solution verification:**
- For \(k = 1\):
\[
a_1 = 1
\]
- For \(k = 2\):
\[
a_2 = 2
\]
- For \(k \geq 3\):
- Evaluate \(a_3\), \(a_4\), and other terms based on the recurrence relation](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa59c8a62-6d27-4c14-9dc5-c4241c4a7fe0%2Fbded28f1-06a8-4d73-8332-4592d23b71df%2F1ew5f8a_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
**(5)** If the series \(\{a_k\}\) satisfies that \(a_1 = 1\), \(a_2 = 2\), and \(a_k - 4a_{k-1} + 3a_{k-2} = 0\) (\(k \geq 3\)), then:
\[ a_k = \frac{1 + \left\lfloor \frac{k-5}{k-6} \right\rfloor}{\left\lfloor \frac{1-5}{1-6} \right\rfloor} \quad (k \geq 1) \]
### Explanation
This problem is related to solving a recurrence relation for the series \(\{a_k\}\).
- The initial conditions are given as \(a_1 = 1\) and \(a_2 = 2\).
- The recurrence relation is \(a_k - 4a_{k-1} + 3a_{k-2} = 0\) for \(k \geq 3\).
The provided formula \(a_k = \frac{1 + \left\lfloor \frac{k-5}{k-6} \right\rfloor}{\left\lfloor \frac{1-5}{1-6} \right\rfloor} \quad (k \geq 1)\) seems to involve floor functions. However, if intended to represent a general solution to the recurrence relation, the floor functions may simplify further.
### Detailed Steps:
1. **Initialize the series:**
- \( a_1 = 1 \)
- \( a_2 = 2 \)
2. **Recurrence relation:**
\[
a_k = 4a_{k-1} - 3a_{k-2} \quad \text{for} \quad k \geq 3
\]
3. **Solution verification:**
- For \(k = 1\):
\[
a_1 = 1
\]
- For \(k = 2\):
\[
a_2 = 2
\]
- For \(k \geq 3\):
- Evaluate \(a_3\), \(a_4\), and other terms based on the recurrence relation
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