If the coefficients of `x^-2` and `x^-4` the expansion of `(x^(1/3) +1/(2x^(1/3)))^18`, are `m` and `n` respectively, then `m/n` is equal to

Options

A. `(5)/(4)`
B. `(4)/(5)`
C. 27
D. 182

Explanation

To solve the problem, we need to find the coefficients of \( x^{-2} \) and \( x^{-4} \) in the expansion of \( \left( x^{1/3} + \frac{1}{2x^{1/3}} \right)^{18} \) and then calculate the ratio \( \frac{m}{n} \).

### Step-by-Step Solution:

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

Thus, the general term becomes:
\[ T_{r+1} = \binom{18}{r} \left( x^{1/3} \right)^{18-r} \left( \frac{1}{2x^{1/3}} \right)^r \]
Simplifying this gives:
\[ T_{r+1} = \binom{18}{r} \left( x^{1/3} \right)^{18-r} \cdot \frac{1}{2^r} \cdot \left( x^{-1/3} \right)^r \]
\[ = \binom{18}{r} \cdot \frac{1}{2^r} \cdot x^{\frac{18-r-r}{3}} = \binom{18}{r} \cdot \frac{1}{2^r} \cdot x^{\frac{18-2r}{3}} \]

2. **Finding Coefficient \( m \) for \( x^{-2} \)**:
We need to find \( r \) such that:
\[ \frac{18 - 2r}{3} = -2 \]
Multiplying both sides by 3:
\[ 18 - 2r = -6 \implies 2r = 24 \implies r = 12 \]
Now, substituting \( r = 12 \) into the general term:
\[ m = \binom{18}{12} \cdot \frac{1}{2^{12}} \]

3. **Finding Coefficient \( n \) for \( x^{-4} \)**:
We need to find \( r \) such that:
\[ \frac{18 - 2r}{3} = -4 \]
Multiplying both sides by 3:
\[ 18 - 2r = -12 \implies 2r = 30 \implies r = 15 \]
Now, substituting \( r = 15 \) into the general term:
\[ n = \binom{18}{15} \cdot \frac{1}{2^{15}} \]

4. **Calculating the Ratio \( \frac{m}{n} \)**:
Now we can find the ratio:
\[ \frac{m}{n} = \frac{\binom{18}{12} \cdot \frac{1}{2^{12}}}{\binom{18}{15} \cdot \frac{1}{2^{15}}} \]
This simplifies to:
\[ \frac{m}{n} = \frac{\binom{18}{12}}{\binom{18}{15}} \cdot 2^{15 - 12} = \frac{\binom{18}{12}}{\binom{18}{15}} \cdot 2^3 \]
Using the property of binomial coefficients:
\[ \binom{18}{12} = \binom{18}{6} \quad \text{and} \quad \binom{18}{15} = \binom{18}{3} \]
Thus:
\[ \frac{m}{n} = \frac{\binom{18}{6}}{\binom{18}{3}} \cdot 8 \]
We know:
\[ \binom{18}{6} = \frac{18!}{6! \cdot 12!} \quad \text{and} \quad \binom{18}{3} = \frac{18!}{3! \cdot 15!} \]
Therefore:
\[ \frac{m}{n} = \frac{3! \cdot 15!}{6! \cdot 12!} \cdot 8 = \frac{6 \cdot 15!}{720 \cdot 12!} \cdot 8 \]
Simplifying gives:
\[ \frac{m}{n} = \frac{6 \cdot 8}{720} \cdot \frac{15 \cdot 14 \cdot 13}{1} = 182 \]

### Final Answer:
\[ \frac{m}{n} = 182 \]