Problem 46 - College Algebra

Question

Find the term indicated in each expansion. \(\left(x+\frac{1}{2}\right)^{8} ;\) fourth term

Step-by-Step Solution

Answer

The fourth term of the binomial \( \left(x+\frac{1}{2}\right)^{8} \) is \( 7x^{5} \).

Step 1: Identify the Values

Here, 'n', which is the power, is 8 and we are asked to find the 4th term, so 'r' is 4. 'x' is 'x' and 'y' is \( \frac{1}{2} \). By substituting these values to the Binomial theorem, the problem becomes a calculation.

Step 2: Calculate the Combination

Calculate the combination for \( {n \choose r-1} = {8 \choose 4-1} = {8 \choose 3} = \frac{8!}{3!(8-3)!} = 56 \)

Step 3: Calculate the Power of 'x'

Next calculate the power of 'x'. it is \( x^{n-r+1}=x^{8-4+1}=x^{5} \)

Step 4: Calculate the Power of 'y'

Finally calculate the power of 'y'. it is \( (\frac{1}{2})^{r-1}=\left(\frac{1}{2}\right)^{4-1}=\left(\frac{1}{2}\right)^{3}=\frac{1}{8} \)

Step 5: Multiply the Outputs

Multiply the binomial coefficient, the power of 'x' and the power of 'y'. This gives \( 56 \cdot x^{5} \cdot \frac{1}{8}=7x^{5} \).

Key Concepts

Combination Calculation

In the Binomial Theorem, a combination is essential for determining which coefficients to apply to each term in the expansion. This process involves using a specific formula to select the number of ways we can pick items from a set without regard to order. For our problem, we calculated the combination using:

• The formula: \({n \choose k} = \frac{n!}{k!(n-k)!}\)
• Where 'n' is the total number of terms (here 8) and 'k' is the term number you are going to minus 1 (here 3 because we want the fourth term).

In this specific case, we calculated \({8 \choose 3}\), which simplifies to 56. This number, 56, is termed the binomial coefficient in this context and acts as a multiplier for the subsequent calculations in the theorem.
Understanding the use of combinations helps solidify the workings of the Binomial Theorem. It gives structure and clarity, confirming that we are working systematically under rules that govern these selections.

Power of a Term

The next phase involves calculating the powers of each component in the binomial. Specifically, we need to determine how the variable \(x\) and the constant term \(\frac{1}{2}\) are raised to certain powers in the context of the term expansion. Given the fourth term's requirements, for \(x\), we derive:

• \(x^{n-r+1} = x^{8-4+1} = x^{5}\)

This operation ensures that \(x\), the primary variable, receives a power that diminishes progressively with each term in the expansion.
Additionally, for the constant term \(\frac{1}{2}\), the power assigned is:

• \(\left(\frac{1}{2}\right)^{r-1} = \left(\frac{1}{2}\right)^{3} = \frac{1}{8}\)

The power of the constant term also follows a decreasing pattern but starts from the inner portion moving outward. This balance of powers of variables and constants assists in retaining the key structure that underpins the Binomial Theorem.

Binomial Coefficient

The binomial coefficient is a crucial element within the Binomial Theorem's framework. It essentially serves as a scalar value multiplying each term in the binomial expansion. In our problem, we determined that this coefficient was 56, significant because it influences the weight of each component term.
The formula we employed, \({n \choose r-1}\), derived from factorial calculations, showcases its inherent dependence on the selection process.
Here’s a breakdown:

• The coefficient 56 derived from \({8 \choose 3}\) is a vital component because it acts as the primary multiplier.
• Once multiplied by the powers of \(x\) and \(\frac{1}{2}\), it results in the term's final value, 7\(x^5\).

By grasping how the coefficient combines with each power, one understands how the complexity of binomials simplifies into coherent, singular terms. This seamless combination is what allows for the expansion of expressions efficiently and correctly.