The 13 th term in the expansion of `(9x-1/(3sqrt(x)))^(18),x gt0` is
(i) `""^(18)C_(12)x^(3)`
(ii) `""^(18)C_(12)x^(6)`
(iii) `""^(18)C_(12)1/(x^(6))`
(iv) `""^(18)C_(12)`

Options

A. `""^(18)C_(12)x^(3)`
B. `""^(18)C_(6)x^(6)`
C. `""^(18)C_(12)1/(x^(6))`
D. `""^(18)C_(6)`

Answer

To find the 13th term in the expansion of \((9x - \frac{1}{3\sqrt{x}})^{18}\), we can use the Binomial Theorem. The general term \(T_{r+1}\) in the expansion of \((a + b)^n\) is given by:

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

In this case, we have:
- \(a = 9x\)
- \(b = -\frac{1}{3\sqrt{x}}\)
- \(n = 18\)

We need to find the 13th term, which corresponds to \(r = 12\) (since \(T_{r+1} = T_{13}\)).

### Step 1: Write the expression for the 13th term
Using the formula for the general term, we have:

\[ T_{13} = \binom{18}{12} (9x)^{18-12} \left(-\frac{1}{3\sqrt{x}}\right)^{12} \]

### Step 2: Simplify the expression
Now, substituting the values:

\[ T_{13} = \binom{18}{12} (9x)^{6} \left(-\frac{1}{3\sqrt{x}}\right)^{12} \]

### Step 3: Calculate \((9x)^6\)
Calculating \((9x)^6\):

\[ (9x)^6 = 9^6 x^6 \]

### Step 4: Calculate \(\left(-\frac{1}{3\sqrt{x}}\right)^{12}\)
Calculating \(\left(-\frac{1}{3\sqrt{x}}\right)^{12}\):

\[ \left(-\frac{1}{3\sqrt{x}}\right)^{12} = \frac{(-1)^{12}}{3^{12} (\sqrt{x})^{12}} = \frac{1}{3^{12} x^6} \]

### Step 5: Combine the results
Now, substituting back into the expression for \(T_{13}\):

\[ T_{13} = \binom{18}{12} \cdot 9^6 x^6 \cdot \frac{1}{3^{12} x^6} \]

### Step 6: Simplify further
The \(x^6\) terms cancel out:

\[ T_{13} = \binom{18}{12} \cdot \frac{9^6}{3^{12}} \]

### Step 7: Simplify \( \frac{9^6}{3^{12}} \)
Since \(9 = 3^2\), we have:

\[ 9^6 = (3^2)^6 = 3^{12} \]

Thus:

\[ \frac{9^6}{3^{12}} = \frac{3^{12}}{3^{12}} = 1 \]

### Final Result
Therefore, we have:

\[ T_{13} = \binom{18}{12} \cdot 1 = \binom{18}{12} \]

So, the 13th term in the expansion is:

\[ \binom{18}{12} \]

### Conclusion
The answer is option (iv) \(\binom{18}{12}\).