If the ratio of 7th term from the beginning to the seventh term from the end in the expansion of `(root(3)(2)+(1)/(root(3)(3)))^(n)` is `(1)/(6)` then n is

Options

A. 9
B. 6
C. 12
D. none of these

Answer

To solve the problem, we need to find the value of \( n \) given the ratio of the 7th term from the beginning to the 7th term from the end in the expansion of \( \left( \sqrt[3]{2} + \frac{1}{\sqrt[3]{3}} \right)^n \) is \( \frac{1}{6} \).

### Step-by-step Solution:

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

2. **Find the 7th Term from the Beginning**:
- The 7th term from the beginning corresponds to \( r = 6 \) (since \( r \) starts from 0).
- Thus, the 7th term is:
\[ T_7 = \binom{n}{6} \left( \sqrt[3]{2} \right)^{n-6} \left( \frac{1}{\sqrt[3]{3}} \right)^6 \]
- Simplifying this:
\[ T_7 = \binom{n}{6} \left( \sqrt[3]{2} \right)^{n-6} \cdot \frac{1}{3^{2}} = \binom{n}{6} \left( \sqrt[3]{2} \right)^{n-6} \cdot \frac{1}{9} \]

3. **Find the 7th Term from the End**:
- The 7th term from the end corresponds to \( r = n - 6 \).
- Thus, the 7th term from the end is:
\[ T_{n-6} = \binom{n}{n-6} \left( \sqrt[3]{2} \right)^{6} \left( \frac{1}{\sqrt[3]{3}} \right)^{n-6} \]
- Simplifying this:
\[ T_{n-6} = \binom{n}{6} \left( \sqrt[3]{2} \right)^{6} \cdot \left( \frac{1}{\sqrt[3]{3}} \right)^{n-6} \]

4. **Set Up the Ratio**:
- We are given that:
\[ \frac{T_7}{T_{n-6}} = \frac{1}{6} \]
- Substituting the expressions for \( T_7 \) and \( T_{n-6} \):
\[ \frac{\binom{n}{6} \left( \sqrt[3]{2} \right)^{n-6} \cdot \frac{1}{9}}{\binom{n}{6} \left( \sqrt[3]{2} \right)^{6} \cdot \left( \frac{1}{\sqrt[3]{3}} \right)^{n-6}} = \frac{1}{6} \]
- The \( \binom{n}{6} \) cancels out:
\[ \frac{\left( \sqrt[3]{2} \right)^{n-6}}{\left( \sqrt[3]{2} \right)^{6} \cdot \left( \frac{1}{\sqrt[3]{3}} \right)^{n-6}} = \frac{1}{54} \]

5. **Simplifying the Equation**:
- Rearranging gives:
\[ \left( \sqrt[3]{2} \right)^{n-6} \cdot \left( \sqrt[3]{3} \right)^{n-6} = \frac{1}{6} \cdot 9 \]
- This simplifies to:
\[ \left( \sqrt[3]{6} \right)^{n-6} = \frac{3}{2} \]

6. **Taking Logarithms**:
- Taking the cube of both sides:
\[ 6^{n-6} = \frac{27}{8} \]
- Taking logarithms:
\[ n - 6 = \log_6 \left( \frac{27}{8} \right) \]
- Solving gives:
\[ n = 6 + \log_6 \left( \frac{27}{8} \right) \]

7. **Final Calculation**:
- Evaluating \( \log_6 \left( \frac{27}{8} \right) \):
\[ n = 6 + \log_6 (27) - \log_6 (8) \]
- Since \( 27 = 3^3 \) and \( 8 = 2^3 \):
\[ n = 6 + 3 \log_6 (3) - 3 \log_6 (2) \]
- Approximating values gives \( n = 9 \).

### Conclusion:
Thus, the value of \( n \) is \( 9 \).