If m is the coefficient of 7th term and n is the coefficient of 13th therm in expansion of `(x^(2/3)/3+1/(2x^(1/3)))^18` then value of `(x^(2/3)/3+1/(2x^(1/3)))^18` is

Options

A. `4/9`
B. `5/9`
C. `9/5`
D. `9/4`

Answer

To solve the problem, we need to find the coefficients of the 7th and 13th terms in the expansion of \((\frac{x^{2/3}}{3} + \frac{1}{2x^{1/3}})^{18}\). We will denote the coefficient of the 7th term as \(m\) and the coefficient of the 13th term as \(n\). Finally, we will calculate the value of \((\frac{n}{m})^{1/3}\).

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term \(T_r\) in the binomial expansion of \((a + b)^n\) is given by:
\[ T_r = \binom{n}{r} a^{n-r} b^r \]
Here, \(a = \frac{x^{2/3}}{3}\), \(b = \frac{1}{2x^{1/3}}\), and \(n = 18\).

2. **Find the 7th Term**:
The 7th term corresponds to \(r = 6\) (since the first term is \(r = 0\)):
\[ T_7 = \binom{18}{6} \left(\frac{x^{2/3}}{3}\right)^{12} \left(\frac{1}{2x^{1/3}}\right)^{6} \]
Simplifying this:
\[ T_7 = \binom{18}{6} \cdot \frac{x^{8}}{3^{12}} \cdot \frac{1}{2^6 x^2} \]
\[ T_7 = \binom{18}{6} \cdot \frac{1}{3^{12} \cdot 2^6} \cdot x^{8 - 2} = \binom{18}{6} \cdot \frac{1}{3^{12} \cdot 2^6} \cdot x^6 \]
Thus, the coefficient \(m\) is:
\[ m = \binom{18}{6} \cdot \frac{1}{3^{12} \cdot 2^6} \]

3. **Find the 13th Term**:
The 13th term corresponds to \(r = 12\):
\[ T_{13} = \binom{18}{12} \left(\frac{x^{2/3}}{3}\right)^{6} \left(\frac{1}{2x^{1/3}}\right)^{12} \]
Simplifying this:
\[ T_{13} = \binom{18}{12} \cdot \frac{x^{4}}{3^{6}} \cdot \frac{1}{2^{12} x^{4}} \]
\[ T_{13} = \binom{18}{12} \cdot \frac{1}{3^{6} \cdot 2^{12}} \cdot x^{4 - 4} = \binom{18}{12} \cdot \frac{1}{3^{6} \cdot 2^{12}} \]
Thus, the coefficient \(n\) is:
\[ n = \binom{18}{12} \cdot \frac{1}{3^{6} \cdot 2^{12}} \]

4. **Calculate \(\frac{n}{m}\)**:
\[ \frac{n}{m} = \frac{\binom{18}{12} \cdot \frac{1}{3^{6} \cdot 2^{12}}}{\binom{18}{6} \cdot \frac{1}{3^{12} \cdot 2^{6}}} \]
\[ = \frac{\binom{18}{12}}{\binom{18}{6}} \cdot \frac{3^{12}}{3^{6}} \cdot \frac{2^{6}}{2^{12}} \]
\[ = \frac{\binom{18}{12}}{\binom{18}{6}} \cdot 3^{6} \cdot \frac{1}{2^{6}} \]
Using the property \(\binom{n}{r} = \binom{n}{n-r}\):
\[ = \frac{\binom{18}{6}}{\binom{18}{6}} \cdot \frac{3^{6}}{2^{6}} = \left(\frac{3}{2}\right)^{6} \]

5. **Calculate \((\frac{n}{m})^{1/3}\)**:
\[ \left(\frac{n}{m}\right)^{1/3} = \left(\left(\frac{3}{2}\right)^{6}\right)^{1/3} = \left(\frac{3}{2}\right)^{2} = \frac{9}{4} \]

### Final Answer:
The value of \((\frac{n}{m})^{1/3}\) is \(\frac{9}{4}\).