Problem 24 - College Algebra

Question

For the following exercises, write an explicit formula for each sequence. $$ 0, \frac{1-e^{1}}{1+e^{2}}, \frac{1-e^{2}}{1+e^{3}}, \frac{1-e^{3}}{1+e^{4}}, \frac{1-e^{4}}{1+e^{5}}, \ldots $$

Step-by-Step Solution

Answer

\( a_n = \begin{cases} 0, & \text{if } n = 1 \\ \frac{1-e^{n-1}}{1+e^{n}}, & \text{if } n \geq 2 \end{cases} \)

Step 1: Identify the Pattern

Observe the given sequence terms: - The first term is 0.- The second term is \( \frac{1-e^{1}}{1+e^{2}} \).- The third term is \( \frac{1-e^{2}}{1+e^{3}} \).- The fourth term is \( \frac{1-e^{3}}{1+e^{4}} \).Notice that each term after the first is of the form \( \frac{1-e^{n}}{1+e^{n+1}} \) for terms starting from index 1.

Step 2: Determine the General Term

If we denote the sequence by \( a_n \), we want to find a general formula for \( n \geq 1 \). The formula appears to take the form:\[a_n = \frac{1-e^{n-1}}{1+e^{n}} \text{ for } n \geq 2\]The first term is special and given as 0, so:\[a_1 = 0\]Therefore, the explicit formula for the sequence is:\[a_n = \begin{cases} 0, & \text{if } n = 1 \ \frac{1-e^{n-1}}{1+e^{n}}, & \text{if } n \geq 2 \end{cases}\]

Step 3: Verify the Formula

Verify the formula by plugging in values:- For \( n = 2 \), \( a_2 = \frac{1-e^{1}}{1+e^{2}} \) matches the second term.- For \( n = 3 \), \( a_3 = \frac{1-e^{2}}{1+e^{3}} \) matches the third term.- For \( n = 4 \), \( a_4 = \frac{1-e^{3}}{1+e^{4}} \) matches the fourth term.The pattern holds, confirming the explicit formula.

Key Concepts

Sequence Pattern Recognition

To solve problems related to sequences, the first step is to grasp the pattern connecting the sequence's terms. Understanding the structure not only aids in simplifying complex problems but also helps in predicting future terms. In this particular sequence, we start with the first term as 0. Subsequent terms, however, form more intricate patterns involving exponential expressions. For instance, looking at the second term, it appears as \( \frac{1-e^{1}}{1+e^{2}} \). This gives a hint that we might be dealing with terms involving fractions and exponents.

As we progress through the sequence, it's evident that each term beyond the first shares a similar structure: \( \frac{1-e^{n-1}}{1+e^{n}} \). Recognizing this repetitive pattern is crucial, as it sets the foundation for determining the general term of the sequence. It's important to carefully compare each term to spot any nuances in their numerators and denominators.

General Term of a Sequence

Once we've deciphered the pattern, our next move is to translate it into a general term. The general term is essentially a formula that represents any term in the sequence based on its position. For the sequence in question, writing a formula involves understanding the structure we identified.

The notation \( a_n \) is commonly used to express a sequence's terms, where \( n \) is the position of the term. Notably, the first term was simply 0. From the second term onwards, we observed a consistent pattern that can be represented as: \[a_n = \begin{cases} 0, & \text{if } n = 1 \ \frac{1-e^{n-1}}{1+e^{n}}, & \text{if } n \geq 2 \end{cases}\]\ The beauty of having a general term is that it provides a direct way to find any term in the sequence without listing all prior terms.

Verification of Sequence Formula

Confidence in your derived formula is essential, and that's where verification comes into play. Verifying the formula involves checking if it correctly represents the given terms in the sequence.

Let's test a few values to ensure the formula's accuracy:

• For \( n = 2 \), compute \( a_2 = \frac{1-e^{1}}{1+e^{2}} \). This matches the second term of the sequence.
• For \( n = 3 \), check \( a_3 = \frac{1-e^{2}}{1+e^{3}} \). It aligns with the third term.
• For \( n = 4 \), see \( a_4 = \frac{1-e^{3}}{1+e^{4}} \), confirming it matches the fourth term.

The accuracy of these computations verifies our formula. This step is crucial as it provides reassurance that the pattern has been captured correctly, allowing for reliable use in predicting further terms.