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 \((2x + \frac{1}{3x})^6\), we will use the Binomial Theorem.

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term \(T_{r+1}\) in the expansion of \((a + b)^n\) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
Here, \(a = 2x\), \(b = \frac{1}{3x}\), and \(n = 6\).

2. **Write the General Term**:
Substituting the values, we have:
\[ T_{r+1} = \binom{6}{r} (2x)^{6-r} \left(\frac{1}{3x}\right)^r \]
This simplifies to:
\[ T_{r+1} = \binom{6}{r} (2^{6-r}) (x^{6-r}) \left(\frac{1}{3^r}\right) (x^{-r}) \]
Combining the powers of \(x\):
\[ T_{r+1} = \binom{6}{r} \cdot 2^{6-r} \cdot \frac{1}{3^r} \cdot x^{6-r-r} = \binom{6}{r} \cdot 2^{6-r} \cdot \frac{1}{3^r} \cdot x^{6-2r} \]

3. **Find the Condition for Independence from \(x\)**:
For the term to be independent of \(x\), the exponent of \(x\) must be zero:
\[ 6 - 2r = 0 \]
Solving for \(r\):
\[ 2r = 6 \implies r = 3 \]

4. **Substitute \(r\) into the General Term**:
Now, we substitute \(r = 3\) into the general term:
\[ T_{4} = \binom{6}{3} \cdot 2^{6-3} \cdot \frac{1}{3^3} \]

5. **Calculate Each Component**:
- Calculate \(\binom{6}{3}\):
\[ \binom{6}{3} = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]
- Calculate \(2^{6-3} = 2^3 = 8\).
- Calculate \(\frac{1}{3^3} = \frac{1}{27}\).

6. **Combine Everything**:
Now, substitute these values back into the term:
\[ T_{4} = 20 \cdot 8 \cdot \frac{1}{27} = \frac{160}{27} \]

### Final Answer:
The term independent of \(x\) in the expansion of \((2x + \frac{1}{3x})^6\) is \(\frac{160}{27}\).