The term independent of x in the expansion of `(2x+1/(3x))^(6)` is

Options

A. 160/9
B. 80/9
C. 160/27
D. 80/3

Answer

To find the term independent of \( x \) in the expansion of \( \left(2x + \frac{1}{3x}\right)^6 \), we will use the Binomial Theorem, which states that the general term in the expansion of \( (a + b)^n \) is given by:

\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]

### Step 1: Identify \( a \), \( b \), and \( n \)

In our case, we have:
- \( a = 2x \)
- \( b = \frac{1}{3x} \)
- \( n = 6 \)

### Step 2: Write the general term

The general term \( T_{r+1} \) in the expansion is:

\[ T_{r+1} = \binom{6}{r} (2x)^{6-r} \left(\frac{1}{3x}\right)^r \]

### Step 3: Simplify the general term

Now, we can simplify this term:

\[ T_{r+1} = \binom{6}{r} (2^{6-r} x^{6-r}) \left(\frac{1}{3^r x^r}\right) \]

Combining the terms gives:

\[ T_{r+1} = \binom{6}{r} \cdot 2^{6-r} \cdot \frac{1}{3^r} \cdot x^{6-r-r} \]

This simplifies to:

\[ T_{r+1} = \binom{6}{r} \cdot 2^{6-r} \cdot \frac{1}{3^r} \cdot x^{6-2r} \]

### Step 4: Find the term independent of \( x \)

For the term to be independent of \( x \), the exponent of \( x \) must be zero:

\[ 6 - 2r = 0 \]

### Step 5: Solve for \( r \)

Solving the equation:

\[ 6 = 2r \implies r = 3 \]

### Step 6: Substitute \( r \) back into the general term

Now, we substitute \( r = 3 \) back into the expression for the general term:

\[ T_{4} = \binom{6}{3} \cdot 2^{6-3} \cdot \frac{1}{3^3} \]

### Step 7: Calculate \( T_{4} \)

Calculating each part:

1. \( \binom{6}{3} = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \)
2. \( 2^{6-3} = 2^3 = 8 \)
3. \( 3^3 = 27 \)

Putting it all together:

\[ T_{4} = 20 \cdot 8 \cdot \frac{1}{27} = \frac{160}{27} \]

### Final Answer

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

\[ \frac{160}{27} \]