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Chapter 11: Problem 14

Find a formula for the general term \(a_{n}\) of the sequence, assuming that the pattern of the first few terms continues. $$ \\{5,1,5,1,5,1, \ldots\\} $$

Short Answer

Expert verified
The formula for the sequence is \(a_n = 5 - 4((n+1) \mod 2)\).

Step by step solution

01

Identify the Pattern

Observe the given sequence: \( \{5,1,5,1,5,1, \ldots\} \). Notice that the sequence alternates between 5 and 1.
02

Determine Periodicity

Since the sequence alternates every two terms, its period is 2. This means every odd-positioned term is 5, and every even-positioned term is 1.
03

Generalize the Pattern

For odd-indexed terms, i.e., when \( n = 1, 3, 5, \ldots \), we have \( a_n = 5 \). For even-indexed terms, i.e., when \( n = 2, 4, 6, \ldots \), we have \( a_n = 1 \).
04

Write the General Formula

To capture this alternation, observe that odd indices can be written as \( 2k - 1 \) and even indices as \( 2k \), for integer \( k \). Thus, the general term \( a_n \) can be defined using the modulo operation:\[a_n = \begin{cases} 5, & \text{if } n \equiv 1 \pmod{2} \1, & \text{if } n \equiv 0 \pmod{2}\end{cases}\]Alternatively, use:\[a_n = 5 - 4((n+1) \mod 2)\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Alternating Sequence
Do you notice something fascinating about the sequence \( \{5, 1, 5, 1, 5, 1, \ldots\} \)? It's an example of an **alternating sequence.** Alternating sequences switch between two or more fixed values following a certain order. In our example, the sequence goes back and forth between 5 and 1. This pattern makes alternating sequences predictable once the pattern's identified. Recognizing the alternating nature is the first step in finding a formula for the general term. Alternating sequences are used in various mathematical calculations and are especially useful in illustrating how different patterns can emerge from simple rules.
Periodicity
The idea of **periodicity** is key to understanding sequences like \( \{5, 1, 5, 1, 5, 1, \ldots\} \). Periodicity refers to something that repeats at regular intervals, like how days of the week cycle every seven days. In a sequence, the period is the number of terms before the sequence pattern repeats.
  • For our sequence, the pattern is \( 5, 1 \), repeating every two terms, so its period is 2.
  • This means after every two entries in the sequence, the pattern starts over.
Recognizing the period in sequences allows you to write a general formula by segmenting the repeating parts, thereby simplifying complex problems into predictable elements.
Even and Odd Indices
Understanding **even and odd indices** is crucial to analyzing sequences like ours. Indices in a sequence refer to the position of terms: first, second, third, etc. In our sequence:
  • Terms at odd indices (1st, 3rd, 5th, etc.) are always 5.
  • Terms at even indices (2nd, 4th, 6th, etc.) are always 1.
This differentiation helps in defining sequences formulaically. By identifying the behavior of terms at even and odd positions, one can effectively characterize the entire sequence and predict any term's value based on its position.
Modulo Operation
The **modulo operation** is a mathematical tool that helps in determining the remainder when one number is divided by another. It's used here to define rules for our sequence:In the given sequence, terms swap between 5 and 1 based on whether their position is odd or even. Using modulo:
  • If \( n \equiv 1 \pmod{2} \), the position \( n \) is odd, and \( a_n = 5 \).
  • If \( n \equiv 0 \pmod{2} \), the position \( n \) is even, and \( a_n = 1 \).
Modulo operation simplifies the task of distinguishing between odd and even indices, enabling a generalized formula like \( a_n = 5 - 4((n+1) \mod 2) \). This formula efficiently captures the entire pattern in one neat expression.

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