Problem 48
Question
For the following exercises, graph the first five terms of the indicated sequence \(a_{n}=\left\\{\begin{array}{ll}\frac{4+n}{2 n} & \text { if } n \text { is even } \\ 3+n & \text { if } n \quad \text { is odd }\end{array}\right.\)
Step-by-Step Solution
Verified Answer
Plot the points on a graph: (1, 4), (2, 1.5), (3, 6), (4, 1), and (5, 8).
1Step 1: Identify the First Five Terms
To graph the first five terms of the sequence, we must first identify what they are. The sequence is defined piecewise, meaning it has two formulas depending on whether the term's position \(n\) is even or odd.
2Step 2: Calculate Terms for Odd n
Apply the odd rule: if \(n\) is odd, use the formula \(3 + n\). For \(n = 1\) (odd), the term is \(3 + 1 = 4\). For \(n = 3\) (odd), the term is \(3 + 3 = 6\). For \(n = 5\) (odd), the term is \(3 + 5 = 8\).
3Step 3: Calculate Terms for Even n
Apply the even rule: if \(n\) is even, use the formula \(\frac{4 + n}{2n}\). For \(n = 2\) (even), the term is \(\frac{4 + 2}{2 \cdot 2} = \frac{6}{4} = 1.5\). For \(n = 4\) (even), the term is \(\frac{4 + 4}{2 \cdot 4} = \frac{8}{8} = 1\).
4Step 4: List the First Five Terms
The sequence of the first five terms, based on the calculations for odd and even \(n\), is: \(n = 1\), the term is 4; \(n = 2\), the term is 1.5; \(n = 3\), the term is 6; \(n = 4\), the term is 1; and \(n = 5\), the term is 8. Thus, the terms are \(4, 1.5, 6, 1, 8\).
5Step 5: Plot the Terms on a Graph
On a coordinate plane, plot the points: \((1, 4)\), \((2, 1.5)\), \((3, 6)\), \((4, 1)\), and \((5, 8)\). Place each pair \((x, y)\) where \(x\) is the position in the sequence (\(n\)) and \(y\) is the value of the term.
Key Concepts
Piecewise FunctionsPlotting PointsOdd and Even RulesAlgebraic Expressions
Piecewise Functions
Piecewise functions are mathematical functions defined by different expressions within different intervals of their domain. They are incredibly useful for describing situations where a rule changes based on the input value. In this exercise, the sequence is specified using two distinct formulas. Each formula applies depending on whether the position number, \( n \), is odd or even.
- For even \( n \), the sequence uses the formula: \( \frac{4+n}{2n} \).
- For odd \( n \), the formula is: \( 3 + n \).
Plotting Points
Plotting points on a graph involves taking each term from the sequence and placing it on a coordinate plane. Each term is represented as a point with an \( x \)-coordinate equal to its position number and a \( y \)-coordinate equal to its calculated value.
To plot these points effectively, follow this process:
To plot these points effectively, follow this process:
- Determine the \( x \)-coordinate from the position number, \( n \).
- Calculate the \( y \)-value using the appropriate formula for each \( n \).
- Find the point \((x, y)\) on the graph and mark it.
Odd and Even Rules
The distinction between odd and even numbers plays a crucial role in many mathematical problems, including ours. Each mathematical formula of our piecewise function applied for \( n \) is based on whether the number is odd or even.
- Odd numbers: These are numbers like 1, 3, 5, etc., which are not divisible by 2 without a remainder.
- Even numbers: These are numbers like 2, 4, 6, etc., divisible by 2.
- When \( n \) is odd, the rule is \( 3 + n \).
- When \( n \) is even, the rule is \( \frac{4+n}{2n} \).
Algebraic Expressions
Algebraic expressions are combinations of numbers, variables, and mathematical operators like addition or division. These expressions form the building blocks of our sequence as they define each term's value based on the position number \( n \).
In this exercise, we see two different algebraic expressions depending on whether \( n \) is even or odd.
In this exercise, we see two different algebraic expressions depending on whether \( n \) is even or odd.
- The expression \( 3 + n \) gives the term value for odd \( n \): simple and straightforward due to its linear nature.
- The expression \( \frac{4+n}{2n} \) appears for even \( n \): this involves both addition and division, reflecting a more complex relational structure.
Other exercises in this chapter
Problem 48
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