Problem 9 - Calculus: Early Transcendentals

Question

List the first five terms of the sequence. \( a_1 = 1, a_{n+1} = 5a_n - 3 \)

Step-by-Step Solution

Answer

The first five terms are 1, 2, 7, 32, and 157.

Step 1: Identify the Given Information

We are given the first term of the sequence, which is \( a_1 = 1 \). The recursive formula provided is \( a_{n+1} = 5a_n - 3 \). Using this formula, we can determine subsequent terms.

Step 2: Find the Second Term

Use the recursive formula to find the second term. Substitute \( n = 1 \) in the formula: \( a_2 = 5a_1 - 3 = 5(1) - 3 = 5 - 3 = 2 \). So, the second term \( a_2 \) is \( 2 \).

Step 3: Find the Third Term

Next, to find the third term, use the formula for \( n = 2 \): \( a_3 = 5a_2 - 3 = 5(2) - 3 = 10 - 3 = 7 \). Therefore, \( a_3 = 7 \).

Step 4: Find the Fourth Term

Now calculate the fourth term. Substitute \( n = 3 \) into the formula: \( a_4 = 5a_3 - 3 = 5(7) - 3 = 35 - 3 = 32 \). Thus, \( a_4 = 32 \).

Step 5: Find the Fifth Term

Finally, find the fifth term by substituting \( n = 4 \): \( a_5 = 5a_4 - 3 = 5(32) - 3 = 160 - 3 = 157 \). So, \( a_5 = 157 \).

Key Concepts

Sequence Terms

When dealing with sequences, we analyze each number in the list as a term of the sequence. In the given problem, we want to find the first five terms of a sequence, which involves using both the initial term and a process to determine subsequent terms.
The starting point of any sequence is critical, and here, the initial term given is \( a_1 = 1 \). Once we have the first term, generating the sequence means repeatedly applying a particular rule to get from one term to the next. This problem uses recursion to find the next terms.
With recursion, we simply take the previous term and use our formula, allowing us to compute all subsequent sequence terms. In this way, we get the terms:

• Second term: \( a_2 = 2 \)
• Third term: \( a_3 = 7 \)
• Fourth term: \( a_4 = 32 \)
• Fifth term: \( a_5 = 157 \)

By understanding each term's relationship through recursion, you gain a clear grasp of how numbers in sequences evolve through rules.

Recursive Formula

Recursive formulas allow us to determine entries of a sequence using the prior terms. It provides an efficient tool for generating terms and helps avoid trial-and-error approaches.
In the given problem, the recursive formula is \( a_{n+1} = 5a_n - 3 \). This formula specifically tells us how to calculate the next term in the sequence from the current term.
Let's break down the formula:

• The term \( a_{n+1} \) indicates the next term in the sequence.
• The expression \( 5a_n \) shows how we modify the current term \( a_n \) by multiplying it by 5.
• The subtraction of 3 at the end ( \( -3 \) ) accounts for a consistent reduction in each step moving forward in the sequence.

By employing this recursive approach, each sequence term can be quickly calculated from its predecessor, leading to a smooth and effective generation of the sequence.

Step by Step Solution

Creating a sequence using a recursive formula involves clearly following specific steps that lead from a simple starting point to the subsequent terms. Here's the breakdown:
**Step 1: Initial Setup**
Identify and write down the first term, which is given as \( a_1 = 1 \). This initial step sets the stage for recursion-based operations.
**Step 2: Applying the Recursive Formula**
For each subsequent term, apply the recursive formula \( a_{n+1} = 5a_n - 3 \) to the most recently determined term. This means:

• Second term, \( a_2 \): Apply formula to \( a_1 \): \( a_2 = 5 \times 1 - 3 = 2 \)
• Third term, \( a_3 \): Apply it to \( a_2 \): \( a_3 = 5 \times 2 - 3 = 7 \)
• Fourth term, \( a_4 \): Apply it to \( a_3 \): \( a_4 = 5 \times 7 - 3 = 32 \)
• Fifth term, \( a_5 \): Apply it to \( a_4 \): \( a_5 = 5 \times 32 - 3 = 157 \)

This step-by-step solution helps illustrate precisely how mathematical operations apply the recursive formula sequentially to reach each subsequent term. It not only ensures accuracy but also builds a methodical understanding of recursive processes in sequences.