Given the recursive sequence ak = (12-06-1) ² - 1 ak-1 -k² + 1 for all k ≥ 1, and a₁ = -4, find a₂ and as-
Given the recursive sequence ak = (12-06-1) ² - 1 ak-1 -k² + 1 for all k ≥ 1, and a₁ = -4, find a₂ and as-
Chapter12: Sequences, Series And Binomial Theorem
Section: Chapter Questions
Problem 328PT: Find the first term and common difference of an arithmetic sequence whose ninth term is -1 and the...
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![### Educational Exercise: Recursive Sequences and Summation
**Instructions:**
Find the following. **SHOW AND JUSTIFY YOUR WORK.**
#### (a) Recursive Sequence Calculation
Given the recursive sequence \( a_k = \left(\frac{1}{2} a_{k-1}\right)^2 - k^2 + 1 \) for all \( k \geq 1 \), and \( a_1 = -4 \), find \( a_2 \) and \( a_3 \).
#### (b) Summation Problems
Compute the following sums using formulas from class. **Do not try to compute something like \(2^{24}\). You may leave your answer in terms of big numbers like this.**
**i.** \(-23 + (-23)^2 + (-23)^3 + \cdots + (-23)^{11}\)
**ii.** \(6 + 12 + 18 + 24 + 30 + \cdots + 1200\)
### Solution Outline:
#### (a) Solving the Recursive Sequence
1. **Find \( a_2 \):**
- Use the given formula with \( k = 2 \).
2. **Find \( a_3 \):**
- Use the same formula with \( k = 3 \), after calculating \( a_2 \).
#### (b) Summation Problems
1. **i. Sum of Powers of \(-23\):**
- Recognize that this is a geometric series with \( a = -23 \) and \( r = -23 \).
2. **ii. Sum of Arithmetic Series:**
- Identify \( a = 6 \) (first term), \( l = 1200 \) (last term), and common difference \( d = 6 \).
### Detailed Solution Steps
- **Recursive Sequence Calculation (a):**
1. **Find \( a_2 \):**
\[
a_2 = \left( \frac{1}{2} a_1 \right)^2 - 2^2 + 1 = \left( \frac{1}{2} \times -4 \right)^2 - 4 + 1 = 4 - 4 + 1 = 1
\]
2. **Find \( a_3 \):**
\[
a_3 = \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffe7646cd-9dad-44ff-a49d-6799ce85d2db%2F13839685-1ea7-411e-8a3d-211861f79ba1%2Fvyeo5vn_processed.png&w=3840&q=75)
Transcribed Image Text:### Educational Exercise: Recursive Sequences and Summation
**Instructions:**
Find the following. **SHOW AND JUSTIFY YOUR WORK.**
#### (a) Recursive Sequence Calculation
Given the recursive sequence \( a_k = \left(\frac{1}{2} a_{k-1}\right)^2 - k^2 + 1 \) for all \( k \geq 1 \), and \( a_1 = -4 \), find \( a_2 \) and \( a_3 \).
#### (b) Summation Problems
Compute the following sums using formulas from class. **Do not try to compute something like \(2^{24}\). You may leave your answer in terms of big numbers like this.**
**i.** \(-23 + (-23)^2 + (-23)^3 + \cdots + (-23)^{11}\)
**ii.** \(6 + 12 + 18 + 24 + 30 + \cdots + 1200\)
### Solution Outline:
#### (a) Solving the Recursive Sequence
1. **Find \( a_2 \):**
- Use the given formula with \( k = 2 \).
2. **Find \( a_3 \):**
- Use the same formula with \( k = 3 \), after calculating \( a_2 \).
#### (b) Summation Problems
1. **i. Sum of Powers of \(-23\):**
- Recognize that this is a geometric series with \( a = -23 \) and \( r = -23 \).
2. **ii. Sum of Arithmetic Series:**
- Identify \( a = 6 \) (first term), \( l = 1200 \) (last term), and common difference \( d = 6 \).
### Detailed Solution Steps
- **Recursive Sequence Calculation (a):**
1. **Find \( a_2 \):**
\[
a_2 = \left( \frac{1}{2} a_1 \right)^2 - 2^2 + 1 = \left( \frac{1}{2} \times -4 \right)^2 - 4 + 1 = 4 - 4 + 1 = 1
\]
2. **Find \( a_3 \):**
\[
a_3 = \
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