Problem 16 - Precalculus Mathematics for Calculus

Question

Find the first five terms of the given recursively defined sequence. $$a_{n}=\frac{1}{1+a_{n-1}} \text { and } a_{1}=1$$

Step-by-Step Solution

Answer

The first five terms are 1, \( \frac{1}{2} \), \( \frac{2}{3} \), \( \frac{3}{5} \), \( \frac{5}{8} \).

Step 1: Understand the recursive formula

The sequence is defined recursively, which means that each term depends on the previous term. The formula for the sequence is given as \( a_n = \frac{1}{1 + a_{n-1}} \). We start with \( a_1 = 1 \).

Step 2: Calculate the second term \(a_2\)

To find \( a_2 \), use the recursive formula with \( a_1 = 1 \): \[ a_2 = \frac{1}{1 + a_1} = \frac{1}{1 + 1} = \frac{1}{2} \]

Step 3: Calculate the third term \(a_3\)

Using the recursive formula and \( a_2 = \frac{1}{2} \), we find: \[ a_3 = \frac{1}{1 + a_2} = \frac{1}{1 + \frac{1}{2}} = \frac{1}{\frac{3}{2}} = \frac{2}{3} \]

Step 4: Calculate the fourth term \(a_4\)

With \( a_3 = \frac{2}{3} \), apply the recursive formula: \[ a_4 = \frac{1}{1 + a_3} = \frac{1}{1 + \frac{2}{3}} = \frac{1}{\frac{5}{3}} = \frac{3}{5} \]

Step 5: Calculate the fifth term \(a_5\)

Using \( a_4 = \frac{3}{5} \), we find: \[ a_5 = \frac{1}{1 + a_4} = \frac{1}{1 + \frac{3}{5}} = \frac{1}{\frac{8}{5}} = \frac{5}{8} \]

Step 6: Summarize the first five terms

Now that we have calculated the terms, the first five terms of the sequence are: 1. \( a_1 = 1 \) 2. \( a_2 = \frac{1}{2} \) 3. \( a_3 = \frac{2}{3} \) 4. \( a_4 = \frac{3}{5} \) 5. \( a_5 = \frac{5}{8} \)

Key Concepts

Sequence Terms

To truly understand recursive sequences, it's essential to grasp the concept of sequence terms. In any sequence, a term is simply an individual element or number that is part of the series.
For our specific exercise, we start with the first term, often referred to as the initial or seed term, which is explicitly given:

• The sequence starts with \( a_1 = 1 \). This term doesn't rely on any other term because it's the foundation or starting point of our sequence.
• The following terms are then derived based on this first term using the recursive formula.

In our recursive sequence, each term is dictated by its position and by the terms before it, which is why identifying the initial term is crucial.
Always remember, each sequence term can unlock the path to more complex sequences and patterns, so understanding them one by one is key.

Recursive Formula

A recursive formula is like a rulebook that defines the relationship between consecutive terms in a sequence. Rather than providing a direct way to calculate any term, it tells you how to derive the next term from the previous one. The formula used in this exercise is:

• \( a_n = \frac{1}{1 + a_{n-1}} \)
• It specifies that to find the \( n^{th} \) term, you use the \((n-1)^{th}\) term.

This formula recursively depends on the output of the previous term to calculate the next term. It’s like building a tower where each new block depends directly on the stability of the block before it.
The neat part about recursive sequences is that they allow us to define potentially complex constructs in a few simple steps. Better yet, understanding the recursive formula gives you the toolkit to unravel any term from the sequence with ease.

Sequence Calculation

Calculating sequence terms for a recursively defined sequence is a step-by-step process. You’ll essentially use the recursive formula iteratively to find each subsequent term. Here's how you approach it:

• Start from what you know: the initial term. In our sequence, this is \( a_1 = 1 \).
• Apply the recursive formula repeatedly for each new term.

For example, to find the second term \(a_2\): \[ a_2 = \frac{1}{1 + a_1} = \frac{1}{2} \] Then, find the third term \(a_3\): \[ a_3 = \frac{1}{1 + a_2} = \frac{2}{3} \] Continue the process methodically in the same manner for each subsequent term.
This calculation mechanic ensures accuracy and is a systematic approach to sequence derivation. With practice in breaking down each calculation, the process becomes less about solving and more about understanding the unfolding pattern.