Problem 11 - Precalculus: Mathematics for Calculus

Question

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

Step-by-Step Solution

Answer

The first four terms are 0, 2, 0, 2, and the 100th term is 2.

Step 1: Understand the Formula

The sequence we are interested in has the formula for the nth term as \( a_{n} = 1 + (-1)^{n} \). This formula means that for each term \( n \), we calculate \( (-1)^{n} \) and add 1 to it.

Step 2: Calculate the First Term

Let's find the first term, where \( n = 1 \). Substitute \( n = 1 \) in the formula: \[a_{1} = 1 + (-1)^{1} = 1 - 1 = 0\]So, the first term is 0.

Step 3: Calculate the Second Term

Now calculate the second term, where \( n = 2 \). Substitute \( n = 2 \) in the formula: \[a_{2} = 1 + (-1)^{2} = 1 + 1 = 2\]So, the second term is 2.

Step 4: Calculate the Third Term

To find the third term, set \( n = 3 \). Substitute \( n = 3 \) in the formula: \[a_{3} = 1 + (-1)^{3} = 1 - 1 = 0\]So, the third term is 0.

Step 5: Calculate the Fourth Term

Determine the fourth term by setting \( n = 4 \). Substitute \( n = 4 \) in the formula: \[a_{4} = 1 + (-1)^{4} = 1 + 1 = 2\]So, the fourth term is 2.

Step 6: Calculate the 100th Term

Finally, find the 100th term, where \( n = 100 \). Substitute \( n = 100 \) in the formula: \[a_{100} = 1 + (-1)^{100} = 1 + 1 = 2\]So, the 100th term is 2.

Key Concepts

Understanding the nth Term Formula

The nth term formula provides a mathematical method to find any term in a sequence without needing to write out all preceding terms. In our sequence, the nth term is defined as \( a_{n} = 1 + (-1)^{n} \). This formula generates terms based on the power of \((-1)\):

• When \( n \) is odd, \((-1)^{n} = -1 \), so \( a_{n} = 1 - 1 = 0 \)
• When \( n \) is even, \((-1)^{n} = 1 \), so \( a_{n} = 1 + 1 = 2 \)

This alternating behavior creates a pattern where the sequence's terms oscillate between 0 and 2, beginning at 0 for \( n = 1 \). Recognizing this behavior is crucial as it simplifies identifying terms at larger values of \( n \).
Whenever you have a sequence defined by a formula, it is beneficial to analyze it to predict outcomes without manual calculations for each term.

Recognizing an Alternating Sequence

In mathematics, an alternating sequence is one where the terms switch or alternate between certain values based on their position in the sequence. Here, our sequence alternates between 0 and 2. This switching happens as a result of the \((-1)^{n}\) term in the formula:

• If \( n \) is odd, the term evaluates to 0.
• If \( n \) is even, the term evaluates to 2.

This alternation is straightforward due to the structure of the function \( (-1)^{n} \). Patterns like these are helpful because they allow us to quickly assess the sequence's properties without computing each term individually.
Alternating sequences naturally simplify the computation of large terms, such as the 100th term here, by applying the pattern instead of tedious calculations.

Calculation of Terms

To calculate specific terms in a sequence effectively, you substitute the desired term number into the nth term formula. For example, to find the first four terms of this sequence, proceed as follows:

• First Term: Set \( n = 1 \), plug into the formula to get \( a_{1} = 0 \).
• Second Term: Set \( n = 2 \), plug into the formula to get \( a_{2} = 2 \).
• Third Term: Set \( n = 3 \), plug into the formula to get \( a_{3} = 0 \).
• Fourth Term: Set \( n = 4 \), plug into the formula to get \( a_{4} = 2 \).

These steps demonstrate the simplicity of using the formula when the sequence follows a clear rule.
For larger sequences, like determining the 100th term, you continue by using the pattern recognition inherent in the formula. Here, \( n = 100 \) is even, so \( a_{100} = 2 \). This method is efficient and reduces the risk of arithmetic errors when identifying terms in a lengthy sequence. Such a structured approach is key when handling more complex problems in arithmetic sequences.