Problem 50 - Precalculus with Limits

Question

In Exercises 47-62, write an expression for the apparent \(n\)th term of the sequence. (Assume that \( n \) begins with 1.) \( 2, -4, 6, -8, 10 \dots \)

Step-by-Step Solution

Answer

The \(n\)th term of the sequence can be given as: \(a_n = 2n \cdot (-1)^{n+1}\).

Step 1: Identifying the pattern

The first step is to identify the pattern in the sequence. This sequence appears to be increasing in absolute value by 2 with each term while alternating signs. The first term is positive, the second term is negative, the third is positive, and so forth. Hence, the sequence can be expressed as \(2, -4, 6, -8, 10 \dots \) = \(2 \cdot 1, 2 \cdot -2, 2 \cdot 3, 2 \cdot -4, 2 \cdot 5 \dots \).

Step 2: Formulating the \(n\)th term rule

Now we need to find a formula that accomplishes these conditions: it should produce a number equivalent to twice the term number, and the sign should alternate between positive and negative. For the sign to alternate, one can multiply by \((-1)^{n+1}\) or \((-1)^n\) depending on if n starts from 0 or 1. As in the exercise \(n\) starts from 1 and we want first term to be positive and the second to be negative, thus we use \((-1)^{n+1}\). So combining this, we can express the \(n\)th term \(a_n\) as \(a_n\) = \(2n \cdot (-1)^{n+1}\).

Key Concepts

Alternating Sequences

Alternating sequences are exciting because they change signs with each term. In the sequence provided, the numbers switch between positive and negative. This happens at each step, creating a zigzag pattern: positive, negative, positive, negative, and so on.

To understand alternating sequences better, visualize the concept with a see-saw. Just as a see-saw goes up and down, an alternating sequence goes positive and then negative.

In mathematics, there’s a neat trick to represent this behavior using powers of (-1). By multiplying each term by (-1) raised to a power related to the position of the term, we can control the sign. For example:

• If you want the first term to be positive when starting from 1, multiply by (-1)^{n+1}.

This cleverly uses even and odd powers of (-1) to alternate signs, since:

• (-1) raised to an even number is positive
• (-1) raised to an odd number is negative

Nth Term Formula

Finding the nth term formula for a sequence requires identifying the consistent rule that generates each term. In every sequence, there's a specific formula or pattern that you can follow to calculate any term.

For the sequence 2, -4, 6, -8, 10, ..., we observed that each term is twice the number of its position in the sequence. So, if you know the term position (n), you multiply it by 2 to get the absolute value. This part gives us a formula like 2n.

Yet, this isn’t enough. To account for the alternating signs, we must include (-1)^{n+1}. Combining these two expressions, the nth term formula turns out to be:\[ a_n = 2n \cdot (-1)^{n+1} \]

This formula reveals that:

• The value 2n generates the number's magnitude.
• The term (-1)^{n+1} controls the sign.

Pattern Recognition

Pattern recognition is the skill of observing how a sequence of numbers behaves. It is like detective work for math, where you look closely for any repetition or rule that governs the sequence.

Recognizing patterns starts with simple observations:

• How do the numbers change?
• Is there a visible pattern in the addition, subtraction, or multiplication between terms?

In our example, we saw that each term grows by 2 in absolute value and switches sign.

By spotting this pattern, you not only understand the sequence better but also lay the groundwork for writing an nth term formula. In sequences, savvy pattern recognition helps you move from confusing strings of numbers to a clear formula you can use to predict any term.

Developing this skill involves practice and a bit of patience but becomes indispensable when solving problems related to sequences and series. It's all about noticing details, much like spotting differences in a picture puzzle. The more patterns you recognize, the better you become at finding formulas and understanding the structure behind the series.