Problem 4 - Thomas Calculus

Question

Each of Exercises \(1-6\) gives a formula for the \(n\) th term \(a_{n}\) of a sequence \(\left\\{a_{n}\right\\} .\) Find the values of \(a_{1}, a_{2}, a_{3},\) and \(a_{4} .\) $$ a_{n}=2+(-1)^{n} $$

Step-by-Step Solution

Answer

The values are: \(a_{1} = 1\), \(a_{2} = 3\), \(a_{3} = 1\), \(a_{4} = 3\).

Step 1: Find the value of a₁

To find \(a_{1}\), substitute \(n = 1\) into the formula:\[a_{1} = 2 + (-1)^{1} = 2 - 1 = 1\]So, \(a_{1} = 1\).

Step 2: Find the value of a₂

To find \(a_{2}\), substitute \(n = 2\) into the formula:\[a_{2} = 2 + (-1)^{2} = 2 + 1 = 3\]Thus, \(a_{2} = 3\).

Step 3: Find the value of a₃

To find \(a_{3}\), substitute \(n = 3\) into the formula:\[a_{3} = 2 + (-1)^{3} = 2 - 1 = 1\]Therefore, \(a_{3} = 1\).

Step 4: Find the value of a₄

To find \(a_{4}\), substitute \(n = 4\) into the formula:\[a_{4} = 2 + (-1)^{4} = 2 + 1 = 3\]So, \(a_{4} = 3\).

Key Concepts

Term Formula

A term formula in a sequence is like a blueprint used to find any specific term within the sequence. In this exercise, the term formula is given as \( a_{n} = 2 + (-1)^{n} \). This formula is mathematical shorthand for telling you how to compute each term of the sequence based on its position, referred to as \( n \). In our formula, \( 2 \) stands for a constant part of the formula, while \((-1)^n\) introduces variability, depending on whether \( n \) is odd or even.
To use a term formula effectively means to plug in different values of \( n \) to obtain the term you are interested in. This process is essential to understand so that you can figure out each term in a sequence systematically.

Alternating Sequence

An alternating sequence is a type of sequence where the terms alternate between different patterns. In our case, the sequence alternates primarily because of the \((-1)^n\) part of the term formula. This particular detail lets the sequence switch between adding and subtracting 1 from 2, depending on whether \( n \) is odd or even.
If \( n \) is odd, \((-1)^n\) will be \(-1\), or negative; when \( n \) is even, \((-1)^n\) will be \(+1\), or positive. This alteration in signs causes the sequence to jump back and forth in value: for example, from 1 to 3 to 1 to 3, and so on. This pattern is typical for alternating sequences and is significant in many mathematical sequences and series problems.
Understanding this concept helps in anticipating the behavior of the sequence as you determine further terms, providing insight into whether the sequence increases, decreases, or maintains a pattern.

Evaluating Terms

When evaluating terms of a sequence, using the term formula is a direct way to determine specific term values. In our sequence, this involves substituting different integer values of \( n \) into the formula \( a_{n} = 2 + (-1)^{n} \). Let’s break down how you would evaluate the first few terms of this sequence:

• For \( n = 1 \), substitute and calculate: \( a_{1} = 2 + (-1)^1 = 2 - 1 = 1 \).
• For \( n = 2 \), substitute and calculate: \( a_{2} = 2 + (-1)^2 = 2 + 1 = 3 \).
• For \( n = 3 \), substitute and calculate: \( a_{3} = 2 + (-1)^3 = 2 - 1 = 1 \).
• For \( n = 4 \), substitute and calculate: \( a_{4} = 2 + (-1)^4 = 2 + 1 = 3 \).

When you evaluate terms this way, you're essentially working through a simple substitution and arithmetic process. This clarity not only helps in understanding how specific terms are derived but also aids in predicting future terms without relying solely on intuition.