Problem 42 - Precalculus: Mathematics for Calculus

Question

Find the indicated terms in the expansion of the given binomial. The term that does not contain \(x\) in the expansion of \(\left(8 x+\frac{1}{2 x}\right)^{8}\).

Step-by-Step Solution

Answer

The constant term is 17920.

Step 1: Identify the Binomial Components

In the expression \(\left(8x + \frac{1}{2x}\right)^8\), identify the two binomial components. Here, \(a = 8x\) and \(b = \frac{1}{2x}\).

Step 2: Use the Binomial Theorem

The binomial theorem states that \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k}\). In this case, \(n = 8\). We need the term that does not contain \(x\), so we will solve for the appropriate \(k\).

Step 3: Determine the General Term

The general term in the expansion is given by \(T_k = \binom{8}{k} (8x)^{8-k} \left(\frac{1}{2x}\right)^k\). Simplifying this gives \(T_k = \binom{8}{k} \cdot 8^{8-k} \cdot x^{8-k} \cdot \left(\frac{1}{2}\right)^k \cdot x^{-k}\).

Step 4: Combine the Powers of x

Combine the powers of \(x\) in the general term: \(x^{8-k} \cdot x^{-k} = x^{8-2k}\). For the term without \(x\), we set \(8-2k = 0\).

Step 5: Solve for k to Find the Term Without x

Solve the equation \(8 - 2k = 0\). \(8 = 2k\) implies \(k = 4\). This means the term without \(x\) occurs when \(k = 4\).

Step 6: Evaluate the Term for k=4

Substitute \(k = 4\) into the general term to find the constant term: \(T_4 = \binom{8}{4} \cdot 8^{8-4} \cdot \left(\frac{1}{2}\right)^4\).

Step 7: Calculate the Binomial Coefficient and the Powers

Calculate \(\binom{8}{4} = 70\), \(8^{4} = 4096\), and \(\left(\frac{1}{2}\right)^4 = \frac{1}{16}\).

Step 8: Combine the Results to Find the Constant Term

Evaluate \(T_4 = 70 \cdot 4096 \cdot \frac{1}{16}\). Simplifying this gives \(T_4 = 70 \cdot 256 = 17920\).

Step 9: Conclusion

The term in the expansion of \(\left(8x + \frac{1}{2x}\right)^8\) that does not contain \(x\) is 17920.

Key Concepts

Binomial Theorem

The Binomial Theorem is a powerful mathematical tool that allows us to expand expressions of the form \((a + b)^n\) into a series involving terms with binomial coefficients. For any integer \(n\), the theorem is expressed as:

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

This expansion provides a way to systematically calculate each term in the set. In our exercise, we applied this theorem to the expansion of \(\left(8x + \frac{1}{2x}\right)^8\). By doing so, we were able to express the binomial as a series of individual terms that could be further simplified. This method is extremely useful for finding specific terms without dealing with the entire polynomial outcome.

General Term

The General Term in a binomial expansion refers to the specific expression for any term within the series resulting from the binomial theorem. It is given by the formula:

• \(T_k = \binom{n}{k} a^{n-k} b^k\)

In our problem, \(a = 8x\) and \(b = \frac{1}{2x}\) with \(n = 8\). Thus, the general term becomes\(T_k = \binom{8}{k} (8x)^{8-k} \left(\frac{1}{2x}\right)^k\).
This term helps identify any selected term, allowing us to solve for specific questions like determining the term that does not contain \(x\). Here, we needed to find when the power of \(x\) was zero, which informed the subsequent steps.

Binomial Coefficient

The Binomial Coefficient, denoted as \(\binom{n}{k}\), represents the number of ways to select \(k\) items from \(n\) options without regard to order. It is calculated using the formula:

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

In our exercise, \(\binom{8}{4}\) was a crucial part. Evaluating it gives us 70. The coefficient is integral to calculating the magnitude of each term in the expansion. Understanding how to compute and apply these coefficients is key to successfully using the binomial theorem for specific expansions.

Powers of Variables

In binomial expansions, handling the Powers of Variables effectively ties into simplifying each term. When each term is expanded, it typically looks like \(a^m b^n\), where the variables are raised to different powers. In our particular example, the general term was written as:

• \(x^{8-k} \cdot x^{-k} = x^{8-2k}\)

Combining these powers and ensuring that the resultant expression meets the desired condition (in our case, having no \(x\)) can simplify the identification of key terms. Knowledge of how to work with exponents is essential for these calculations, ensuring our steps correctly transition from algebraic expressions to clear numerical results.