Chapter 9: Problem 47
Show that the sequence \(s_{n}\) satisfies the recurrence relation. $$\begin{aligned} &s_{n}=3 n-2\\\ &s_{n}=s_{n-1}+3 \text { for } n>1 \text { and } s_{1}=1 \end{aligned}$$
Short Answer
Expert verified
The sequence satisfies the recurrence relation.
Step by step solution
01
Verify the Initial Condition
The recurrence relation states that for \(n = 1\), \(s_1 = 1\). Let's check if \(s_n = 3n - 2\) holds true for \(n = 1\). Substitute \(n = 1\) into the formula: \[s_1 = 3(1) - 2 = 3 - 2 = 1\]This matches the given initial condition \(s_1 = 1\), so the initial step is verified.
02
Assume the Recurrence is True for \(n = k\)
Assume that the recurrence relation holds for \(n = k\), meaning:\[s_k = 3k - 2\]We will use this assumption in the next step to prove that it holds for \(n = k + 1\).
03
Prove the Recurrence for \(n = k + 1\)
We need to show that if the relation holds for \(n = k\), then it also holds for \(n = k + 1\).The recurrence relation is given as:\[s_{n} = s_{n-1} + 3\]Substitute \(n = k + 1\) and use the assumption:\[s_{k+1} = s_k + 3\]From the assumption, \(s_k = 3k - 2\), so:\[s_{k+1} = (3k - 2) + 3 = 3k + 1\]Now, substitute \(n = k + 1\) into the explicit formula and check:\[s_{k+1} = 3(k + 1) - 2 = 3k + 3 - 2 = 3k + 1\]Since this matches \(s_{k+1} = 3k + 1\), the expression is verified.
04
Conclusion
Since we have verified that \(s_1\) matches the initial condition and confirmed that if the formula holds for \(n = k\), it holds for \(n = k + 1\), by mathematical induction, the sequence \(s_n = 3n - 2\) satisfies the recurrence relation with \(s_{n} = s_{n-1}+3\) for \(n > 1\).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Mathematical Induction
Mathematical induction is a powerful technique often used to prove that a statement, formula, or property holds for all natural numbers. The process resembles a chain reaction, where the validity at step one ensures the validity at the next step and so on.
There are two primary phases in applying mathematical induction:
There are two primary phases in applying mathematical induction:
- Base Case: First, we demonstrate that the statement holds true for the initial value, often denoted as \(n = 1\). In the given sequence \(s_n\), checking the initial condition \(s_1 = 1\) ensures that our base case is valid.
- Inductive Step: Assume that the statement is true for some number \(k\), and then prove it for \(k + 1\). In the exercise, assuming \(s_k = 3k - 2\) and showing it holds for \(s_{k+1}\) establishes the inductive step.
Sequence
A sequence is an ordered list of numbers defined by a specific rule. In mathematics, sequences are one of the fundamental concepts as they allow modeling and solving various problems.
The sequence \(s_n = 3n - 2\) is said to be defined explicitly, meaning each term can be calculated directly without relying on the previous one.
The explicit formula gives us a more straightforward calculation:
The sequence \(s_n = 3n - 2\) is said to be defined explicitly, meaning each term can be calculated directly without relying on the previous one.
The explicit formula gives us a more straightforward calculation:
- For \(n = 1\), as shown: \(s_1 = 3(1) - 2 = 1\).
- For \(n = k + 1\), \(s_{k+1} = 3(k + 1) - 2 = 3k + 3 - 2 = 3k + 1\).
Initial Condition
An initial condition is the starting point for a sequence or recurrence relation. It specifies the value of the first term, laying the foundation for further calculations.
For recurrence relations, the initial condition acts as the anchor point:
For recurrence relations, the initial condition acts as the anchor point:
- In the context of the sequence \(s_n\), the initial condition is \(s_1 = 1\).
- It helps to uniquely determine the terms that follow, based on the rule defined by the recurrence relation \(s_{n} = s_{n-1} + 3\).