Problem 45 - College Algebra

Question

In Exercises 39-48, find the term indicated in each expansion. $$\left(x-\frac{1}{2}\right)^{y} ; \text { fourth term }$$

Step-by-Step Solution

Answer

The fourth term in the expansion of \( \left(x-\frac{1}{2}\right)^{y}\) is \(-\frac{1}{8}C(y, 3) * x^{y-3}\).

Step 1: Identify the Variables in the Binomial Expansion

Here, the binomial is \(x - \frac{1}{2}\) and the exponent, n, is \(y\). The task is to find the fourth term, so \(k = 4\). Thus, the specific variables for this calculation will be \(a = x, b = -\frac{1}{2}\), \(n = y\), and \(k = 4\).

Step 2: Substitute the Variables into the Binomial Theorem

To find the fourth term, substitute the variables into the theorem widely known as binomial theorem to get: \(C(y, 4-1) * x^{y-4+1} * \left(-\frac{1}{2}\right)^{4-1}\).

Step 3: Simplify the Fourth Term

Simplify the expression to get the fourth term: \(C(y, 3) * x^{y-3} * \left(-\frac{1}{8}\right) = -\frac{1}{8}C(y, 3) * x^{y-3}\).

Key Concepts

Binomial Theorem

The binomial theorem is a powerful tool in algebra that helps us expand expressions raised to a power. Specifically, it provides a formula for expanding binomials, which are expressions with two terms like \((x - \frac{1}{2})\), raised to any positive integer power \(n\). The binomial theorem states:

• \((a + b)^n = \sum_{k=0}^{n} C(n, k) \cdot a^{n-k} \cdot b^k\)

Here, \(C(n, k)\) represents a binomial coefficient, which is often read as "n choose k." This coefficient tells us how many ways we can choose \(k\) elements from a set of \(n\) elements. By using this theorem, we can identify any specific term in the expansion, such as the fourth term. Simply determine the position of the term you are interested in by adjusting the value of \(k\). In our exercise, the fourth term involves solving for \(k = 4\). This means we find the third term in the sequence, as the index starts from zero.
This simplification allows you to directly access any particular part of the expanded algebraic expression without expanding it fully.

Combinatorics

Combinatorics is a branch of mathematics dealing with combinations of objects. In the context of the binomial theorem, it helps us understand how the coefficients in binomial expansions are calculated. Combinatorics revolves around two main ideas: permutations and combinations.

• Permutations relate to the arrangement of items, where order matters.
• Combinations focus on the selection of items, where order does not matter.

When using the binomial theorem, the binomial coefficient \(C(n, k)\) is of particular interest. This expression refers to the number of ways to choose \(k\) items from \(n\), and is calculated using the formula:

• \(C(n, k) = \frac{n!}{k!(n-k)!}\)

Here, \(!\) denotes the factorial of a number, which is the product of that number and all the integers below it. For example, \(4! = 4 \times 3 \times 2 \times 1 = 24\). Understanding this concept is essential for determining the coefficients in any expanded binomial expression. This knowledge allows students to identify terms in binomial expansions, such as the fourth term in an expansion like \((x - \frac{1}{2})^y\), by computing \(C(y, 3)\).

Algebra Expansion

Algebra expansion involves expressing an algebraic expression as a sum of terms. This process makes it easier to perform operations and solve equations involving algebraic terms. When expanding binomials using the binomial theorem, each term in the expansion has a specific structure: it combines powers of the individual elements of the binomial.

• The first element in the binomial, raised to decreasing powers.
• The second element, raised to increasing powers, produces a term.
• The binomial coefficient for that term.

For example, to determine the fourth term in the expansion \((x - \frac{1}{2})^y\), we can substitute the relevant values into the generalized term expression:

• \(C(y, 3) \cdot x^{y-3} \cdot \left(-\frac{1}{2}\right)^3\)

Simplifying leads to:

• -\(\frac{1}{8}C(y, 3) \cdot x^{y-3}\)

Though performing algebra expansion can seem challenging, it becomes more manageable with practice and understanding. The process helps in extracting meaningful information from complex expressions, by breaking them down into orderly parts.