Problem 8 - Algebra and Trigonometry

Question

Find the first four terms and the 100th term of the sequence. \(a_{n}=(-1)^{n+1} \frac{n}{n+1}\)

Step-by-Step Solution

Answer

First four terms: \(\frac{1}{2}, -\frac{2}{3}, \frac{3}{4}, -\frac{4}{5}\). 100th term: \(-\frac{100}{101}\).

Step 1: Understanding the Formula for Terms of the Sequence

The formula given for the sequence is \(a_n = (-1)^{n+1} \frac{n}{n+1}\). The sequence is defined using a general term formula where \(n\) denotes the position of the term.

Step 2: Calculating the First Term

To find the first term \(a_1\), substitute \(n = 1\) into the formula. Thus, \(a_1 = (-1)^{1+1} \frac{1}{1+1} = (-1)^{2} \frac{1}{2} = 1 \times \frac{1}{2} = \frac{1}{2}\).

Step 3: Calculating the Second Term

For the second term \(a_2\), substitute \(n = 2\) into the formula. Thus, \(a_2 = (-1)^{2+1} \frac{2}{2+1} = (-1)^{3} \frac{2}{3} = -1 \times \frac{2}{3} = -\frac{2}{3}\).

Step 4: Calculating the Third Term

For the third term \(a_3\), substitute \(n = 3\) into the formula. Thus, \(a_3 = (-1)^{3+1} \frac{3}{3+1} = (-1)^{4} \frac{3}{4} = 1 \times \frac{3}{4} = \frac{3}{4}\).

Step 5: Calculating the Fourth Term

For the fourth term \(a_4\), substitute \(n = 4\) into the formula. Thus, \(a_4 = (-1)^{4+1} \frac{4}{4+1} = (-1)^{5} \frac{4}{5} = -1 \times \frac{4}{5} = -\frac{4}{5}\).

Step 6: Calculating the 100th Term

For the 100th term \(a_{100}\), substitute \(n = 100\) into the formula. Thus, \(a_{100} = (-1)^{100+1} \frac{100}{100+1} = (-1)^{101} \frac{100}{101} = -1 \times \frac{100}{101} = -\frac{100}{101}\).

Key Concepts

Understanding the Sequence Formula

An algebraic sequence can be defined by a specific formula that allows us to calculate any term in the sequence based purely on the position of the term, denoted as \( n \). Consider the sequence given by the formula: \[a_n = (-1)^{n+1} \frac{n}{n+1}. \]This formula does a couple of things:

• The expression \((-1)^{n+1}\) gives us a sign change with each subsequent term. This is essential to introduce the element of an alternating sequence.
• The fraction \(\frac{n}{n+1}\) determines the actual value of each term, increasing with each increase in \( n \).

By substituting different values for \( n \), you can generate as many terms as needed in the sequence. This makes it a powerful tool for exploring and understanding the pattern of the sequence.

Techniques for Calculating Terms

Calculating terms in an algebraic sequence involves substituting the position of the term, denoted as \( n \), into the given formula. Let's break down the process using the sequence formula discussed earlier.
To find any term \( a_n \) in the sequence:

• Start by substituting the desired position \( n \) into the formula.
• Simplify the power \((-1)^{n+1}\) which alternates signs from positive to negative through subsequent terms.
• Calculate the fraction \(\frac{n}{n+1}\) to finalize the term's value.
Examples:
• For \( n = 1 \), \( a_1 = \frac{1}{2} \), the fraction is positive, since \((-1)^2 = 1\).
• For \( n = 100 \), \( a_{100} = -\frac{100}{101} \), note the sign change, as \((-1)^{101} = -1\).
This approach allows you to systematically determine any term within the sequence, providing insight into the sequence's behavior.

Patterns in an Alternating Sequence

An alternating sequence is a sequence where the signs of the terms switch in a regular pattern. In our given sequence, this alternation is driven by the expression \((-1)^{n+1}\). Let's delve into how this impacts the sequence:

• The sequence alternates signs because the power of -1 oscillates between even and odd as \( n \) increments. Odd powers result in negative products, while even powers yield positive products.
• The repetition is every other term, which means every two terms, the sign flips. This results in a predictable wave-like pattern.
• The fraction \(\frac{n}{n+1}\) acts more subtly, increasing the magnitude of the terms but adhering to the alternating sign rule.

Understanding these patterns gives us the tools to predict the nature of future terms without even calculating them fully. Such insight is valuable in making conjectures or predictions for larger sets of terms in algebraic sequences.