https://file.aiquickdraw.com/mathbot/file/1725616974809hyxwvtzg.jpg

We are tasked with finding the term independent of \(x\) in the expansion of:

\[
\frac{1}{x} \left( \frac{1}{3x} - \frac{x}{2} \right)^9
\]

### Step 1: Expansion Setup

Let the expression inside the parentheses be:

\[
\left( \frac{1}{3x} - \frac{x}{2} \right)^9
\]

This can be expanded using the **binomial theorem**:

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

Here:
- \( a = \frac{1}{3x} \)
- \( b = -\frac{x}{2} \)
- \( n = 9 \)

The general term in the binomial expansion is:

\[
T_k = \binom{9}{k} \left( \frac{1}{3x} \right)^{9-k} \left( -\frac{x}{2} \right)^k
\]

### Step 2: Simplifying Each Term

For each term, simplify:

\[
T_k = \binom{9}{k} \left( \frac{1}{3x} \right)^{9-k} \left( -\frac{x}{2} \right)^k
\]

This simplifies to:

\[
T_k = \binom{9}{k} \left( \frac{1}{3^{9-k} x^{9-k}} \right) \left( \frac{(-1)^k x^k}{2^k} \right)
\]

\[
T_k = \binom{9}{k} \frac{(-1)^k}{3^{9-k} 2^k} \cdot \frac{x^k}{x^{9-k}}
\]

\[
T_k = \binom{9}{k} \frac{(-1)^k}{3^{9-k} 2^k} \cdot x^{2k-9}
\]

### Step 3: Multiply by \( \frac{1}{x} \)

Now, multiply the entire expansion by \( \frac{1}{x} \):

\[
T_k' = \frac{1}{x} T_k = \binom{9}{k} \frac{(-1)^k}{3^{9-k} 2^k} \cdot x^{2k-10}
\]

### Step 4: Find Term Independent of \(x\)

For the term to be independent of \(x\), we need:

\[
2k - 10 = 0 \implies k = 5
\]

### Step 5: Substitute \(k = 5\)

Substitute \( k = 5 \) into the expression for \( T_k' \):

\[
T_5' = \binom{9}{5} \frac{(-1)^5}{3^{9-5} 2^5} = \binom{9}{5} \frac{-1}{3^4 2^5}
\]

\[
T_5' = \frac{126}{81 \times 32} \cdot (-1)
\]

\[
T_5' = \frac{-126}{2592} = -\frac{7}{144}
\]

Thus, the term independent of \(x\) in the expansion is:

\[
\boxed{-\frac{7}{144}}
\]

Would you like a more detailed breakdown, or do you have any further questions?

Here are 5 related questions:
1. How do you apply the binomial theorem to complex expansions?
2. What are the conditions for a term to be independent of \(x\) in an expression?
3. How does the sign of terms alternate in binomial expansions with negative terms?
4. How can we calculate binomial coefficients for larger values of \(n\) and \(k\)?
5. What happens if the powers in the binomial expansion involve fractional exponents?

**Tip:** Always double-check the powers of variables in each term when solving for specific conditions (like independence of a variable).

Find the term independent of x in the expansion of (1/x)(1/3x - x/2)^9.

Learn how to find the term independent of x in the expansion of (1/x)(1/3x - x/2)^9 using the binomial theorem. This detailed solution walks through the process step-by-step, covering essential algebraic concepts for Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation