If the 5th term is the term independent of x in the expansion of `(x^(2//3) + 1/x)^n` then n =

Options

A. 10
B. 8
C. 7
D. 12

Answer

To solve the problem, we need to find the value of \( n \) such that the 5th term in the expansion of \( (x^{2/3} + \frac{1}{x})^n \) is independent of \( x \).

### 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, \( a = x^{2/3} \) and \( b = \frac{1}{x} \).

2. **Substituting Values**:
For our case, the general term becomes:
\[ T_{r+1} = \binom{n}{r} (x^{2/3})^{n-r} \left(\frac{1}{x}\right)^r \]
This simplifies to:
\[ T_{r+1} = \binom{n}{r} x^{\frac{2}{3}(n-r)} x^{-r} = \binom{n}{r} x^{\frac{2n}{3} - \frac{2r}{3} - r} \]

3. **Finding the 5th Term**:
The 5th term corresponds to \( r = 4 \) (since \( r+1 = 5 \)):
\[ T_5 = \binom{n}{4} x^{\frac{2n}{3} - \frac{2 \cdot 4}{3} - 4} \]
Thus, we have:
\[ T_5 = \binom{n}{4} x^{\frac{2n}{3} - \frac{8}{3} - 4} \]

4. **Setting the Exponent of \( x \) to Zero**:
For the term to be independent of \( x \), the exponent must equal zero:
\[ \frac{2n}{3} - \frac{8}{3} - 4 = 0 \]

5. **Solving for \( n \)**:
Rearranging the equation:
\[ \frac{2n}{3} - \frac{8}{3} - \frac{12}{3} = 0 \]
\[ \frac{2n - 20}{3} = 0 \]
Multiplying through by 3:
\[ 2n - 20 = 0 \]
\[ 2n = 20 \]
\[ n = 10 \]

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