Problem 39 - College Algebra with Corequisite Support

Question

For the following exercises, find the indicated term of each binomial without fully expanding the binomial. The eighth term of \(\left(\frac{y}{2}+\frac{2}{x}\right)^{9}\)

Step-by-Step Solution

Answer

The eighth term is \( \frac{4608 \, y^2}{x^7} \).

Step 1: Identify the binomial theorem

To find the indicated term in the binomial expansion of \( \left(\frac{y}{2} + \frac{2}{x}\right)^{9} \), we use the binomial theorem. The \(k^{th}\) term in the expansion of \( (a + b)^{n} \) is given by:\[ T_{k+1} = \binom{n}{k} \cdot a^{n-k} \cdot b^{k} \]

Step 2: Set parameters for the eighth term

We need to find the eighth term of the expansion, which corresponds to \(T_8\). The formula gives \( k = 7 \) since \( k+1 = 8 \). Substitute \( n = 9 \), \( a = \frac{y}{2} \), \( b = \frac{2}{x} \), and \( k = 7 \) into the binomial term formula.

Step 3: Calculate the binomial coefficient

The coefficient is \( \binom{9}{7} \). Calculate it using the binomial coefficient formula:\[ \binom{9}{7} = \binom{9}{2} = \frac{9 \times 8}{2 \times 1} = 36 \].

Step 4: Substitute into the formula

Substitute the values into the binomial term formula:\[ T_{8} = \binom{9}{7} \cdot \left(\frac{y}{2}\right)^{9-7} \cdot \left(\frac{2}{x}\right)^{7} \]Simplify to:\[ T_{8} = 36 \cdot \left(\frac{y}{2}\right)^2 \cdot \left(\frac{2}{x}\right)^7 \].

Step 5: Simplify the expression

Simplify \( \left(\frac{y}{2}\right)^{2} \) to \( \frac{y^2}{4} \) and \( \left(\frac{2}{x}\right)^{7} \) to \( \frac{2^7}{x^7} \).Calculate \( 2^{7} = 128 \).Therefore:\[ T_8 = 36 \cdot \frac{y^2}{4} \cdot \frac{128}{x^7} = 36 \cdot \frac{y^2}{4} \cdot \frac{128}{x^7} \]Which simplifies to:\[ T_8 = \frac{36 \, \cdot \, 128 \, \cdot \, y^2}{4x^7} \].

Step 6: Compute the final term

Simplify the expression to get:\[ T_8 = \frac{36 \, \cdot \, 128 \, \cdot \, y^2}{4x^7} = \frac{4608 \, y^2}{x^7} \].

Key Concepts

Binomial Expansion

Binomial expansion is a powerful algebraic tool that expresses the expansion of expressions raised to a power. For two terms, or binomials, in the form of \((a + b)^n\), the Binomial Theorem provides a formula to expand this without having to multiply everything out manually. Each term in the expansion has the components of binomial coefficients, powers of the first term \(a\), and powers of the second term \(b\). This enables us to identify specific terms in the expansion quickly.

The general term in a binomial expansion \( T_{k+1} \) is given as:

• \(T_{k+1} = \binom{n}{k} \cdot a^{n-k} \cdot b^k \)

This formula makes it possible to find specific terms without fully expanding the whole expression. By knowing the position \(k\) of the term you want, and other given values, you can substitute and get the term directly.

Binomial Coefficient

In binomial expansion, each term is multiplied by a special number called a binomial coefficient. This binomial coefficient \(\binom{n}{k}\) is derived from the number of ways to choose \(k\) elements from a set of \(n\) elements, hence also known as a combination. It is a fundamental part of the binomial expansion, ensuring that each term in the expansion is correctly weighted.

The binomial coefficient is calculated by:

• \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\)

where \(!\) denotes factorial, which means multiplying a series of descending natural numbers. For example, \(n! = n \times (n-1) \times ... \times 1\). Knowing how to compute this coefficient is key to finding exact terms in the binomial expansion.

Algebraic Expressions

Algebraic expressions are statements composed of variables and constants, combined using algebraic operations like addition, subtraction, multiplication, and division. In the context of our binomial expansion problem, expressions include the components \(\frac{y}{2}\) and \(\frac{2}{x}\).

Understanding how to manipulate these expressions is crucial when simplifying the terms in binomial expansions. Simplifying an algebraic expression involves:

• Combining like terms
• Using algebraic properties to factor and reduce terms

For instance, to simplify \((\frac{y}{2})^2\), you multiply the expression by itself, resulting in \(\frac{y^2}{4}\). Similarly, applying powers to fractions requires careful attention to both the numerator and denominator.