If the expansion of `(3/7sqrtx-5/2(1)/(xsqrtx))^(13n) xgt0` contains a term independent of x, then n should be a multiple of ______

Answer

To solve the problem, we need to find the value of \( n \) such that the expansion of

\[ \left( \frac{3}{7} \sqrt{x} - \frac{5}{2} \frac{1}{x \sqrt{x}} \right)^{13n} \]

contains a term independent of \( x \).

### Step-by-step Solution:

1. **Identify the General Term**:
The general term in the binomial expansion of \( (a + b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^r b^{n-r} \]
Here, we will replace \( a \) and \( b \) with \( \frac{3}{7} \sqrt{x} \) and \( -\frac{5}{2} \frac{1}{x \sqrt{x}} \) respectively, and \( n \) with \( 13n \).

2. **Substituting Values**:
The general term becomes:
\[ T_{r+1} = \binom{13n}{r} \left( \frac{3}{7} \sqrt{x} \right)^r \left( -\frac{5}{2} \frac{1}{x \sqrt{x}} \right)^{13n - r} \]

3. **Simplifying the General Term**:
This can be simplified as:
\[ T_{r+1} = \binom{13n}{r} \left( \frac{3^r}{7^r} x^{r/2} \right) \left( -\frac{5^{13n - r}}{2^{13n - r} x^{13n - r} \sqrt{x}^{13n - r}} \right) \]
Combining the powers of \( x \):
\[ T_{r+1} = \binom{13n}{r} \left( \frac{3^r (-5)^{13n - r}}{7^r 2^{13n - r}} \right) x^{\frac{r}{2} - (13n - r) - \frac{13n - r}{2}} \]

4. **Finding the Power of \( x \)**:
The exponent of \( x \) simplifies to:
\[ \frac{r}{2} - (13n - r) - \frac{13n - r}{2} = \frac{r}{2} - 13n + r - \frac{13n}{2} + \frac{r}{2} = r - 13n \]

5. **Setting the Exponent to Zero**:
For the term to be independent of \( x \), we set the exponent equal to zero:
\[ r - 13n = 0 \implies r = 13n \]

6. **Finding the Value of \( r \)**:
From the earlier steps, we also derived:
\[ r = \frac{39n}{4} \]

7. **Equating the Two Expressions for \( r \)**:
Set the two expressions for \( r \) equal:
\[ 13n = \frac{39n}{4} \]

8. **Solving for \( n \)**:
Multiply both sides by 4 to eliminate the fraction:
\[ 52n = 39n \implies 52n - 39n = 0 \implies 13n = 0 \]
This implies \( n \) must be a multiple of 4 for \( r \) to be an integer.

### Conclusion:
Thus, \( n \) should be a multiple of **4**.