Problem 33 - Precalculus

Question

For the following exercises, write the first eight terms of the sequence. $$ a_{1}=2, \mathrm{a}_{2}=10, a_{n}=\frac{2\left(a_{n-1}+2\right)}{a_{n-2}} $$

Step-by-Step Solution

Answer

The first eight terms are 2, 10, 12, 2.8, 0.8, 2, 10, 12.

Step 1: Identify Given Values

We are given the first two terms of the sequence: \(a_1 = 2\) and \(a_2 = 10\). The recursive formula for the sequence is \(a_n = \frac{2(a_{n-1} + 2)}{a_{n-2}}\).

Step 2: Calculate the Third Term

Using the formula \(a_3 = \frac{2(a_2 + 2)}{a_1}\), we substitute the known values: \(a_2 = 10\) and \(a_1 = 2\). This gives us \(a_3 = \frac{2(10 + 2)}{2} = \frac{2 \times 12}{2} = 12\).

Step 3: Calculate the Fourth Term

For the fourth term, use \(a_4 = \frac{2(a_3 + 2)}{a_2}\). Substitute \(a_3 = 12\) and \(a_2 = 10\) to get \(a_4 = \frac{2(12 + 2)}{10} = \frac{2 \times 14}{10} = 2.8\).

Step 4: Calculate the Fifth Term

Next, compute \(a_5 = \frac{2(a_4 + 2)}{a_3}\). Substitute \(a_4 = 2.8\) and \(a_3 = 12\) into the formula: \(a_5 = \frac{2(2.8 + 2)}{12} = \frac{2 \times 4.8}{12} = 0.8\).

Step 5: Calculate the Sixth Term

For the sixth term, use \(a_6 = \frac{2(a_5 + 2)}{a_4}\). Substitute \(a_5 = 0.8\) and \(a_4 = 2.8\): \(a_6 = \frac{2(0.8 + 2)}{2.8} = \frac{2 \times 2.8}{2.8} = 2\).

Step 6: Calculate the Seventh Term

To find \(a_7\), use the formula \(a_7 = \frac{2(a_6 + 2)}{a_5}\). Substitute \(a_6 = 2\) and \(a_5 = 0.8\): \(a_7 = \frac{2(2 + 2)}{0.8} = \frac{2 \times 4}{0.8} = 10\).

Step 7: Calculate the Eighth Term

Finally, use \(a_8 = \frac{2(a_7 + 2)}{a_6}\) to calculate the eighth term. Substitute \(a_7 = 10\) and \(a_6 = 2\): \(a_8 = \frac{2(10 + 2)}{2} = \frac{2 \times 12}{2} = 12\).

Key Concepts

Sequence Terms Calculation

When calculating the terms of a recursive sequence, you must use the information you have about the previous terms to find the next ones. This approach begins by identifying any given terms and then methodically applying the recursive formula.
For example, in the sequence given, you start with two known values, often called initial terms. In our current exercise, we have that:

• \( a_1 = 2 \)
• \( a_2 = 10 \)

From here, we calculate subsequent terms using the recursive formula:

Calculating the Next Terms

Consider finding \( a_3 \). You substitute \( a_2 \) and \( a_1 \) into the recursive formula:
\( a_3 = \frac{2(a_2 + 2)}{a_1} \). With \( a_2 = 10 \) and \( a_1 = 2 \), this gives you \( a_3 = 12 \).

Thus, calculating each term involves:

• Inserting the most recent and relevant terms into the formula.
• Performing the necessary arithmetic calculations.

This method requires care to ensure each calculation is based on the correct prior terms.

Recursive Formula

A recursive formula is essential for generating terms in a sequence, relying on preceding terms rather than positioning an index directly.
In mathematical sequences, a recursive formula provides a way to connect each term to one or more of its preceding terms.

How It Works

In the given problem, the recursive formula is:\[ a_n = \frac{2(a_{n-1} + 2)}{a_{n-2}} \]This defines each term, \( a_n \), in terms of \( a_{n-1} \) (the previous term) and \( a_{n-2} \) (the term before the previous term). Key features include:

• The formula makes use of arithmetic operations to transform previous terms.
• It involves both an addition \( (+2) \) and a division \( (\div) \) pattern.

By understanding its structure, you can apply calculations consistently to produce new terms following the predetermined pattern.

Mathematical Sequences

Mathematical sequences are ordered sets of numbers governed by specific rules. Sequences can be defined recursively or explicitly. In recursive sequences, each term is defined relative to one or more previous terms rather than a specific formula based on the term's position.

Key Characteristics

• They start with one or more initial terms.
• A formula is used to determine all subsequent terms.
• Seemingly complex sequences can emerge from simple recursive patterns, displaying fascinating behaviors.

In the exercise, the sequence derives its complexity from its recursive nature, illustrating how previous terms dictate the sequence's growth. This dynamic is typical with recursively defined sequences where initial conditions profoundly influence subsequent values rather than a fixed index position in the sequence.