Problem 30 - College Algebra with Corequisite Support

Question

For the following exercises, write the first five terms of the sequence. \(a_{1}=-30, \quad a_{n}=\left(2+a_{n-1}\right)\left(\frac{1}{2}\right)^{n}\)

Step-by-Step Solution

Answer

The first five terms are -30, -7, -5/8, 11/128, and 267/4096.

Step 1: Identify the first term

The first term of the sequence is given as \(a_1 = -30\). This is our starting point.

Step 2: Calculate the second term

To find the second term \(a_2\), use the formula: \(a_n = \left( 2 + a_{n-1} \right) \left(\frac{1}{2}\right)^n\).Substitute \(n = 2\) and \(a_1 = -30\):\[a_2 = \left( 2 + (-30) \right) \left(\frac{1}{2}\right)^2\]Simplify:\[a_2 = (-28) \cdot \left(\frac{1}{4}\right) = -7\].

Step 3: Calculate the third term

Substitute \(n = 3\) and \(a_2 = -7\):\[a_3 = \left( 2 + (-7) \right) \left(\frac{1}{2}\right)^3\]Simplify:\[a_3 = (-5) \cdot \left(\frac{1}{8}\right) = -\frac{5}{8}\].

Step 4: Calculate the fourth term

Substitute \(n = 4\) and \(a_3 = -\frac{5}{8}\):\[a_4 = \left( 2 + \left(-\frac{5}{8}\right) \right) \left(\frac{1}{2}\right)^4\]Simplify:\[a_4 = \left(\frac{16}{8} - \frac{5}{8}\right) \cdot \frac{1}{16} = \frac{11}{8} \cdot \frac{1}{16} = \frac{11}{128}\].

Step 5: Calculate the fifth term

Substitute \(n = 5\) and \(a_4 = \frac{11}{128}\):\[a_5 = \left( 2 + \frac{11}{128} \right) \left(\frac{1}{2}\right)^5\]Simplify:Convert 2 to a fraction with denominator 128: \(2 = \frac{256}{128}\).\[a_5 = \left(\frac{256}{128} + \frac{11}{128}\right) \cdot \frac{1}{32} = \frac{267}{128} \cdot \frac{1}{32} = \frac{267}{4096}\].

Key Concepts

Recursive Formulas

In mathematics, a recursive formula defines each term in a sequence using previous terms. It's like a baking recipe that calls for the result of the previous step. This kind of formula is essential in sequences because it allows us to find terms without direct calculation, the term-by-term way.
For instance, in the given problem, the recursive formula is: \[a_n = \left(2 + a_{n-1}\right)\left(\frac{1}{2}\right)^n\] Here, each term \(a_n\) is derived from the previous term \(a_{n-1}\) by following the steps defined. The beauty of recursive formulas lies in its systematic process, taking one step at a time to reach the next stage.
To effectively use a recursive formula:

• Always keep track of the previous term (e.g., \(a_{n-1}\)).
• Follow the recursive rule precisely to avoid mistakes.
• Start from a known term, usually the first term, \(a_1\).

This particular sequence is not linear, adding to the complexity by including both addition with the previous term and a multiplying factor that changes with \(n\). Understanding the recursive formula is key to solving such sequence problems.

Arithmetic Sequences

An arithmetic sequence is a type of mathematical sequence where each term after the first is the sum of the previous term and a constant difference. However, in the problem provided, the sequence isn't strictly arithmetic because it doesn't have a constant difference.
Instead, it's a blend of recursive and exponential components. The formula \(a_n = \left(2 + a_{n-1}\right)\left(\frac{1}{2}\right)^n\) might mislead one to think it's an arithmetic model. But remember, true arithmetic sequences have a constant addition or subtraction pattern.
So, while the problem sequence starts with \(a_1 = -30\), the transition to the next terms doesn't follow a straightforward arithmetic difference because of the multiplying factor \(\left(\frac{1}{2}\right)^n\). Unlike arithmetic sequences which are predictable, this sequence requires careful consideration at each step due to the changing factor. For students, it is important not to confuse recursive sequences with arithmetic ones.

First Terms Calculation

Calculating the first few terms of a sequence gives insight into its behavior. With the first term given as \(a_1 = -30\), we can determine the next terms by applying the recursive formula methodically:

• Start with the initial term \(a_1 = -30\).
  
• For the second term \(a_2\), substitute values into the recursive formula: \(a_2 = \left(2 + (-30)\right) \left(\frac{1}{2}\right)^2 = -28 \cdot \frac{1}{4} = -7\).
  
• The third term \(a_3\) involves \(a_2\) and follows: \(a_3 = \left(2 + (-7)\right) \left(\frac{1}{2}\right)^3 = -5 \cdot \frac{1}{8} = -\frac{5}{8}\).
  
• Proceed to the fourth term \(a_4 = \left(2 + \left(-\frac{5}{8}\right)\right) \left(\frac{1}{2}\right)^4 = \frac{11}{128}\).
  
• Finally, for the fifth term \(a_5 = \left(2 + \frac{11}{128}\right) \left(\frac{1}{2}\right)^5 = \frac{267}{4096}\).
  

Each calculation relies on substituting the previous term into the formula to derive its value. By understanding and practicing these calculations, students gain proficiency in handling sequences beyond mere memorization.