If the 6th term from the beginning is equal to the 6th term from the end in the expansion of `(2^(1//5)+1/(3^(1//5)))^(n)`, then n is equal to:

Options

A. 7
B. 9
C. 10
D. 12

Answer

To solve the problem, we need to find the value of \( n \) such that the 6th term from the beginning is equal to the 6th term from the end in the expansion of \( \left( 2^{1/5} + \frac{1}{3^{1/5}} \right)^n \).

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term \( T_r \) in the expansion of \( (x + y)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} x^r y^{n-r} \]
Here, \( x = 2^{1/5} \) and \( y = \frac{1}{3^{1/5}} \).

2. **Find the 6th Term from the Beginning**:
The 6th term from the beginning corresponds to \( r = 5 \):
\[ T_6 = \binom{n}{5} \left( 2^{1/5} \right)^5 \left( \frac{1}{3^{1/5}} \right)^{n-5} \]
Simplifying this, we get:
\[ T_6 = \binom{n}{5} \cdot 2 \cdot \frac{1}{3^{(n-5)/5}} = \binom{n}{5} \cdot 2 \cdot 3^{-(n-5)/5} \]

3. **Find the 6th Term from the End**:
The 6th term from the end corresponds to \( r = n - 5 \):
\[ T_{n-5} = \binom{n}{n-5} \left( 2^{1/5} \right)^{n-5} \left( \frac{1}{3^{1/5}} \right)^5 \]
This simplifies to:
\[ T_{n-5} = \binom{n}{5} \cdot \left( 2^{1/5} \right)^{n-5} \cdot \frac{1}{3} = \binom{n}{5} \cdot 2^{(n-5)/5} \cdot \frac{1}{3} \]

4. **Set the Two Terms Equal**:
We set \( T_6 \) equal to \( T_{n-5} \):
\[ \binom{n}{5} \cdot 2 \cdot 3^{-(n-5)/5} = \binom{n}{5} \cdot 2^{(n-5)/5} \cdot \frac{1}{3} \]
Canceling \( \binom{n}{5} \) from both sides (assuming \( n \geq 5 \)):
\[ 2 \cdot 3^{-(n-5)/5} = 2^{(n-5)/5} \cdot \frac{1}{3} \]

5. **Rearranging the Equation**:
Multiply both sides by \( 3^{(n-5)/5} \):
\[ 2 \cdot 3^{(n-5)/5} = 2^{(n-5)/5} \]

6. **Equate the Powers**:
Since the bases are different, we can equate the powers:
\[ 2^{1} = 2^{(n-5)/5} \quad \text{and} \quad 3^{(n-5)/5} = 3^{1} \]
This gives us two equations:
\[ \frac{n-5}{5} = 1 \quad \text{(for base 2)} \]
\[ \frac{n-5}{5} = 1 \quad \text{(for base 3)} \]

7. **Solve for \( n \)**:
From either equation:
\[ n - 5 = 5 \implies n = 10 \]

### Conclusion:
Thus, the value of \( n \) is \( \boxed{10} \).