https://file.aiquickdraw.com/mathbot/file/1730027888224befygnli.jpg

We are asked to determine the possible values of \(n\) such that the expansion of \((8x^3 - \frac{1}{2x})^n\) has a non-zero **constant term**. 

Let’s walk through the problem:

### Step 1: General Binomial Expansion Formula
The binomial expansion of \((a + b)^n\) is given by:

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

For our case:
- \(a = 8x^3\)
- \(b = -\frac{1}{2x}\)

### Step 2: Identifying the Constant Term
In the expansion, each term will be of the form:

\[
\binom{n}{k} (8x^3)^{n-k} \left(-\frac{1}{2x}\right)^k
\]

Simplify the powers of \(x\) for each term:
- \((8x^3)^{n-k} = 8^{n-k} \cdot x^{3(n-k)}\)
- \(\left(-\frac{1}{2x}\right)^k = (-1)^k \cdot \frac{1^k}{2^k} \cdot x^{-k}\)

Thus, a general term in the expansion becomes:

\[
\binom{n}{k} \cdot 8^{n-k} \cdot (-1)^k \cdot \frac{1}{2^k} \cdot x^{3(n-k) - k}
\]

### Step 3: Condition for a Constant Term
For a constant term, the power of \(x\) must be zero. That is:

\[
3(n - k) - k = 0
\]

Solve for \(k\):

\[
3n - 3k - k = 0 \quad \Rightarrow \quad 3n - 4k = 0 \quad \Rightarrow \quad 3n = 4k \quad \Rightarrow \quad k = \frac{3n}{4}
\]

### Step 4: Ensuring \(k\) is an Integer
Since \(k\) must be an integer (as it’s a binomial coefficient index), \(3n\) must be divisible by 4. This means \(n\) must be a multiple of 4. So, the possible values of \(n\) are:

\[
n = 4, 8, 12, \dots
\]

That is, \(n \in \{4m \mid m \in \mathbb{Z}^+\}\).

### Final Answer
The expansion has a non-zero constant term for all positive integers \(n\) that are multiples of 4, i.e., \(n = 4, 8, 12, \dots\).

---

Do you need further details or a specific step clarified?  
Here are five related questions for exploration:
1. How would the solution change if the term inside was \((ax^2 - \frac{b}{x})^n\)?
2. How does the binomial expansion change when \(n\) is not an integer?
3. How do we handle binomial expansions when negative or fractional exponents are involved?
4. Can this problem be solved using generating functions?
5. How is the binomial theorem connected to combinatorics?

**Tip:** In binomial expansions, always check the powers carefully when looking for specific terms like constants or coefficients!

Consider the expansion of \((8x^3 - \frac{1}{2x})^n\) where \(n \in \mathbb{Z}^+\). Determine all possible values of \(n\) for which the expansion has a non-zero constant term.

Explore how to determine all possible values of n for which the expansion of (8x^3 - 1/(2x))^n has a non-zero constant term. This problem uses the binomial theorem and algebraic manipulation, suitable for Grades 11-12 students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation