The term independent of `'x'` in the expansion of `(9x-(1)/(3sqrt(x)))^(18)`, `x gt 0` , is `alpha` times the corresponding binomial coefficient. Then `'alpha'` is

Options

A. `3`
B. `(1)/(3)`
C. `-(1)/(3)`
D. `1`

Explanation

To find the term independent of \( x \) in the expansion of \( (9x - \frac{1}{3\sqrt{x}})^{18} \), we will follow these steps:

### Step 1: Identify the general term in the binomial expansion
The general term \( T_r \) in the expansion of \( (a + b)^n \) is given by:
\[ T_r = \binom{n}{r} a^{n-r} b^r \]
In our case, \( a = 9x \), \( b = -\frac{1}{3\sqrt{x}} \), and \( n = 18 \). Therefore, the general term can be expressed as:
\[ T_r = \binom{18}{r} (9x)^{18-r} \left(-\frac{1}{3\sqrt{x}}\right)^r \]

### Step 2: Simplify the general term
Substituting the values of \( a \) and \( b \):
\[ T_r = \binom{18}{r} (9^{18-r} x^{18-r}) \left(-\frac{1}{3}\right)^r \left(\frac{1}{\sqrt{x}}\right)^r \]
This simplifies to:
\[ T_r = \binom{18}{r} 9^{18-r} \left(-\frac{1}{3}\right)^r x^{18-r - \frac{r}{2}} \]
\[ = \binom{18}{r} 9^{18-r} \left(-\frac{1}{3}\right)^r x^{18 - r - \frac{r}{2}} \]
\[ = \binom{18}{r} 9^{18-r} \left(-\frac{1}{3}\right)^r x^{18 - \frac{3r}{2}} \]

### Step 3: Find the term independent of \( x \)
For the term to be independent of \( x \), the exponent of \( x \) must be zero:
\[ 18 - \frac{3r}{2} = 0 \]
Solving for \( r \):
\[ \frac{3r}{2} = 18 \implies 3r = 36 \implies r = 12 \]

### Step 4: Substitute \( r \) back into the general term
Now, we substitute \( r = 12 \) into the general term:
\[ T_{12} = \binom{18}{12} 9^{18-12} \left(-\frac{1}{3}\right)^{12} x^{18 - \frac{3 \cdot 12}{2}} \]
\[ = \binom{18}{12} 9^6 \left(-\frac{1}{3}\right)^{12} \]

### Step 5: Calculate the coefficient
Calculating the coefficient:
\[ = \binom{18}{12} 9^6 \cdot \left(-\frac{1}{3^{12}}\right) \]
\[ = \binom{18}{12} \cdot \frac{9^6}{3^{12}} \cdot (-1) \]
Since \( 9 = 3^2 \):
\[ = \binom{18}{12} \cdot \frac{(3^2)^6}{3^{12}} \cdot (-1) \]
\[ = \binom{18}{12} \cdot \frac{3^{12}}{3^{12}} \cdot (-1) \]
\[ = \binom{18}{12} \cdot (-1) \]

### Step 6: Relate to \( \alpha \)
According to the problem, the term independent of \( x \) is \( \alpha \) times the corresponding binomial coefficient:
\[ \alpha \cdot \binom{18}{12} = -\binom{18}{12} \]
Thus, we find:
\[ \alpha = -1 \]

### Final Answer
The value of \( \alpha \) is:
\[ \alpha = -1 \]