Problem 28 - College Algebra

Question

For the following exercises, write the first five terms of the sequence. $$ a_{1}=-4, a_{n}=\frac{a_{n-1}+2 n}{a_{n-1}-1} $$

Step-by-Step Solution

Answer

The first five terms are -4, 0, -6, \(-\frac{2}{7}\), and \(-\frac{68}{9}\).

Step 1: Calculate the 2nd Term

To find the second term of the sequence, we use the given recursive formula. Substitute \( n = 2 \) and \( a_1 = -4 \) into the formula: \[ a_2 = \frac{a_1 + 2 \times 2}{a_1 - 1} = \frac{-4 + 4}{-4 - 1} = \frac{0}{-5} = 0 \]So, \( a_2 = 0 \).

Step 2: Calculate the 3rd Term

Now, using \( a_2 = 0 \), substitute \( n = 3 \) into the recursive formula: \[ a_3 = \frac{a_2 + 2 \times 3}{a_2 - 1} = \frac{0 + 6}{0 - 1} = \frac{6}{-1} = -6 \]So, \( a_3 = -6 \).

Step 3: Calculate the 4th Term

Using \( a_3 = -6 \), substitute \( n = 4 \) into the recursive formula: \[ a_4 = \frac{a_3 + 2 \times 4}{a_3 - 1} = \frac{-6 + 8}{-6 - 1} = \frac{2}{-7} = -\frac{2}{7} \]So, \( a_4 = -\frac{2}{7} \).

Step 4: Calculate the 5th Term

Finally, using \( a_4 = -\frac{2}{7} \), substitute \( n = 5 \) into the formula: \[ a_5 = \frac{a_4 + 2 \times 5}{a_4 - 1} = \frac{-\frac{2}{7} + 10}{-\frac{2}{7} - 1} \]First, simplify the expression:\[ a_5 = \frac{-\frac{2}{7} + \frac{70}{7}}{-\frac{2}{7} - \frac{7}{7}} = \frac{\frac{68}{7}}{-\frac{9}{7}} = \frac{68}{7} \times -\frac{7}{9} = -\frac{68}{9} \]Therefore, \( a_5 = -\frac{68}{9} \).

Key Concepts

Algebraic Expressions

Algebraic expressions are the language of algebra, used to represent numbers, variables, and operations in a concise way. In a recursive sequence like the one in our exercise, an algebraic expression is used to define the relation between the terms. For example, in the expression given for the sequence:
\( a_{n} = \frac{a_{n-1} + 2n}{a_{n-1} - 1} \), we can see how each term depends on the previous term \( a_{n-1} \), and also incorporates the index \( n \).
Understanding the structure of an algebraic expression involves recognizing the components like numbers (coefficients), operations (addition, subtraction, multiplication, division), and variables (like n in our case). These expressions can perform complex calculations succinctly, allowing the easy manipulation of formulas as shown when substituting values within the sequence's recursive formula.
In practical terms, when dealing with algebraic expressions in sequences, observe:

• How each part of the expression interacts with others; for instance, how terms can cancel out as seen in the calculation of \( a_2 \).
• How changes in variables () affect the outcome, reinforcing the importance of understanding each element.

Gaining proficiency with algebraic expressions enhances one's ability to decode and solve mathematical problems.

Term Calculation

Term calculation involves substituting given values into an algebraic formula to arrive at specific results. This is particularly important in finding terms in sequences, whether they are arithmetic, geometric, or recursive like in our exercise. Each step in our sequence calculation entails computing the next term by using the outcome of previous terms.
Calculating terms can involve various arithmetic operations, as seen where each term \( a_n \) is determined from \( a_{n-1} \) by doing arithmetic involving operations on both the sequence terms and their indices. Hence, when solving sequences:

• Always carefully substitute the previous term value and adjust the expression according to the position (n).
• Keep track of each operation to prevent errors, especially in recursive sequences where one mistake can cascade through the entire problem solution.

Additionally, recognizing patterns in sequence calculations can also simplify the work and enable predictions about future terms.

Fraction Simplification

Simplifying fractions is a key part of working with sequences in algebra, especially when terms are expressed as fractions, as in our given problem. Fraction simplification involves reducing a fraction to its simplest form, making it easier to understand and use in further calculations.
Fractions appear frequently when dividing terms, as shown when solving for \( a_5 \). Simplification requires:

• Identifying common factors in the numerator and denominator, which can cancel each other out.
• Understanding how to manipulate fractions that include whole numbers and fractions, such as turning mixed numbers into improper fractions when necessary.

In our exercises, we often factor out numbers or change forms to simplify, like converting \(-\frac{2}{7} + 10\) to a common denominator to ease calculation.
Simplifying fractions not only makes math easier but reduces complexities, ensuring any arithmetic is efficiently performed, which is crucial in both manual calculations and when coding these formulas into a computer program.