Problem 9 - College Algebra with Modeling and Visualization

Question

Find the first four terms of the sequence. \(a_{n}=(-1)^{n-1}\left(\frac{2^{n}}{1+2^{n}}\right)\)

Step-by-Step Solution

Answer

The first four terms are \( \frac{2}{3}, -\frac{4}{5}, \frac{8}{9}, -\frac{16}{17} \).

Step 1: Substitute n = 1 into the formula

Start by calculating the first term of the sequence, which is when \( n = 1 \). Substitute 1 for \( n \) in the given formula: \[ a_1 = (-1)^{1-1} \left(\frac{2^1}{1+2^1}\right) \] Simplify the expression to find \( a_1 \).

Step 2: Simplify for n = 1

Calculate the powers and simplify: \[ a_1 = (-1)^0 \left(\frac{2}{1+2}\right) = 1 \times \frac{2}{3} = \frac{2}{3} \] The first term \( a_1 \) is \( \frac{2}{3} \).

Step 3: Substitute n = 2 into the formula

Next, compute the second term by setting \( n = 2 \) in the formula: \[ a_2 = (-1)^{2-1} \left(\frac{2^2}{1+2^2}\right) \] Simplify the expression for \( a_2 \).

Step 4: Simplify for n = 2

Calculate: \[ a_2 = (-1)^1 \left(\frac{4}{1+4}\right) = -1 \times \frac{4}{5} = -\frac{4}{5} \] Therefore, the second term \( a_2 \) is \( -\frac{4}{5} \).

Step 5: Substitute n = 3 into the formula

Find the third term by substituting \( n = 3 \) in the formula: \[ a_3 = (-1)^{3-1} \left(\frac{2^3}{1+2^3}\right) \] Simplify this expression for \( a_3 \).

Step 6: Simplify for n = 3

Calculate: \[ a_3 = (-1)^2 \left(\frac{8}{1+8}\right) = 1 \times \frac{8}{9} = \frac{8}{9} \] Hence, \( a_3 \) is \( \frac{8}{9} \).

Step 7: Substitute n = 4 into the formula

Now compute the fourth term by setting \( n = 4 \) in the sequence: \[ a_4 = (-1)^{4-1} \left(\frac{2^4}{1+2^4}\right) \] Simplify to find \( a_4 \).

Step 8: Simplify for n = 4

Calculate: \[ a_4 = (-1)^3 \left(\frac{16}{1+16}\right) = -1 \times \frac{16}{17} = -\frac{16}{17} \] Therefore, \( a_4 \) is \( -\frac{16}{17} \).

Key Concepts

Algebra

Algebra is a branch of mathematics that uses symbols and letters to represent numbers and quantities in equations and formulas. In the context of sequences, algebraic expressions enable us to define rules for generating terms. These rules are usually given in the form of an equation or a formula.

• Variables: In sequences, variables like \( n \) are often used as placeholders for terms' positions in the sequence.
• Operations: Algebra uses operations like addition, subtraction, multiplication, division, and exponentiation to form expressions.

Algebraic manipulation involves substituting values for variables, performing operations according to the order of operations, and simplifying expressions. Using algebra, we can derive terms based on a given rule, as seen when calculating each term in the sequence from the exercise.
Understanding these algebraic principles is crucial for handling sequences, as it allows you to compute specific terms without listing all previous ones.

Terms of a Sequence

A sequence is an ordered list of numbers, where each number is a term. The sequence in our exercise is defined using a formula that incorporates algebraic expressions.

• First Term: Often denoted as \( a_1 \), it sets the foundation for the sequence.
• Subsequent Terms: These are calculated by plugging values into the sequence's formula.

To find each term:

• Identify the position \( n \) of the term you want to calculate.
• Substitute \( n \) into the sequence formula to find the value of the term.

Having the ability to calculate terms individually proves handy, especially in more complex sequences where direct listing is impractical. Each term is unique based on the position it occupies in the sequence and reflects how sequences can alternate or stay constant, as seen in the given formula with alternating signs.

Recursive Formula

A recursive formula defines each term in a sequence using the preceding terms. However, in the exercise given, the formula to find each term is not recursive but explicit.

• Explicit Formula: Allows calculation of any term directly, without needing previous terms.
• Recursive Example: A recursive formula for sequences typically looks like \( a_n = a_{n-1} + d \), where \( d \) is the common difference.

In comparison to recursive formulas, explicit ones provide a direct computation method as evident in the exercise solution. In an explicit formula like \( a_{n}=(-1)^{n-1}\left(\frac{2^{n}}{1+2^{n}}\right) \), each term is calculated independently by substituting \( n \) directly into the formula.
Understanding the difference between recursive and explicit formulas helps us better decide which approach is appropriate for generating terms of a sequence effectively.