Solve the first-order linear recurrence relation: Sn+1 = 5 Sn + 1, with S0=1. You may use the general solution given on P.342.
Solve the first-order linear recurrence relation: Sn+1 = 5 Sn + 1, with S0=1. You may use the general solution given on P.342.
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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Solve the first-order linear recurrence relation: Sn+1 = 5 Sn + 1, with S0=1. You may use the general solution given on P.342.
![The given text is drawn from a chapter on sequences and series, explaining solutions for specific sequences:
The formula for \( S_n \) is divided into two cases:
- If \( a = 1 \),
\[
S_n = I + nc \quad \text{for} \ n \in \mathbb{N};
\]
- If \( a \neq 1 \),
\[
S_n = a^nA + \frac{c}{1 - a} \quad \text{for} \ n \in \mathbb{N}.
\]
When \( a = 1 \), any particular solution is obtained by determining a specific numerical value for \( I \). It's defined for a particular entry \( S_j \), solving the equation:
\[
J = I + jc \quad \Rightarrow \quad I = J - jc \quad \text{since} \ S_j = I + jc.
\]
A particular solution may have \( I = 0 \).
When \( a \neq 1 \), a particular solution is determined by \( A \):
\[
J = Aa^j + \frac{c}{1 - a} \quad \text{for} \ A
\]
which gives:
\[
A = \frac{1}{a^j} \left[J - \frac{c}{1 - a}\right].
\]
A particular solution may have \( A = 0 \).
**Example 8.2.1: The Towers of Hanoi**
The recurrence equation for the number of moves in the Towers of Hanoi is a first-order linear recurrence equation:
\[
T_n = 2T_{n-1} + 1.
\]
With \( a = 2 \) and \( c = 1 \), we have:
\[
\frac{c}{1-a} = \frac{1}{1-2} = -1,
\]
resulting in the formula:
\[
T_n = 2^n[I - (-1)] + (-1) = 2^n[I + 1] - 1.
\]
Assuming \( T \) has domain \( \mathbb{N} \) and denoting \( T_0 \) by \( I \), several particular solutions are given:
1. If \( I = 0 \), \( T = \{0, 1,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa4792360-22b4-40c0-a48a-02ce03c9f6f7%2F157a57ad-46f0-4bde-8935-6337adb201e9%2F57aysob_processed.png&w=3840&q=75)
Transcribed Image Text:The given text is drawn from a chapter on sequences and series, explaining solutions for specific sequences:
The formula for \( S_n \) is divided into two cases:
- If \( a = 1 \),
\[
S_n = I + nc \quad \text{for} \ n \in \mathbb{N};
\]
- If \( a \neq 1 \),
\[
S_n = a^nA + \frac{c}{1 - a} \quad \text{for} \ n \in \mathbb{N}.
\]
When \( a = 1 \), any particular solution is obtained by determining a specific numerical value for \( I \). It's defined for a particular entry \( S_j \), solving the equation:
\[
J = I + jc \quad \Rightarrow \quad I = J - jc \quad \text{since} \ S_j = I + jc.
\]
A particular solution may have \( I = 0 \).
When \( a \neq 1 \), a particular solution is determined by \( A \):
\[
J = Aa^j + \frac{c}{1 - a} \quad \text{for} \ A
\]
which gives:
\[
A = \frac{1}{a^j} \left[J - \frac{c}{1 - a}\right].
\]
A particular solution may have \( A = 0 \).
**Example 8.2.1: The Towers of Hanoi**
The recurrence equation for the number of moves in the Towers of Hanoi is a first-order linear recurrence equation:
\[
T_n = 2T_{n-1} + 1.
\]
With \( a = 2 \) and \( c = 1 \), we have:
\[
\frac{c}{1-a} = \frac{1}{1-2} = -1,
\]
resulting in the formula:
\[
T_n = 2^n[I - (-1)] + (-1) = 2^n[I + 1] - 1.
\]
Assuming \( T \) has domain \( \mathbb{N} \) and denoting \( T_0 \) by \( I \), several particular solutions are given:
1. If \( I = 0 \), \( T = \{0, 1,
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