For a positive integer n, if the expanison of
`(5/x^(2) + x^(4))` has a term independent of x, then n can be

Options

A. 18
B. 27
C. 36
D. 45

Explanation

To find the positive integer \( n \) such that the expansion of \( \left( \frac{5}{x^2} + x^4 \right)^n \) has a term independent of \( x \), we can follow these steps:

### Step 1: Identify the general term in the binomial expansion
The general term \( T_{r+1} \) in the expansion of \( \left( \frac{5}{x^2} + x^4 \right)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} \left( \frac{5}{x^2} \right)^{n-r} \left( x^4 \right)^r \]

### Step 2: Simplify the general term
Simplifying the term, we have:
\[ T_{r+1} = \binom{n}{r} \cdot 5^{n-r} \cdot \frac{1}{x^{2(n-r)}} \cdot x^{4r} \]
This can be rewritten as:
\[ T_{r+1} = \binom{n}{r} \cdot 5^{n-r} \cdot x^{4r - 2(n-r)} \]
\[ = \binom{n}{r} \cdot 5^{n-r} \cdot x^{4r - 2n + 2r} \]
\[ = \binom{n}{r} \cdot 5^{n-r} \cdot x^{6r - 2n} \]

### Step 3: Set the exponent of \( x \) to zero
For the term to be independent of \( x \), the exponent of \( x \) must be zero:
\[ 6r - 2n = 0 \]

### Step 4: Solve for \( n \) in terms of \( r \)
Rearranging the equation gives:
\[ 2n = 6r \quad \Rightarrow \quad n = 3r \]

### Step 5: Determine possible values of \( n \)
Since \( n \) must be a positive integer, \( r \) can take any positive integer value. Therefore, the possible values of \( n \) are:
- If \( r = 1 \), then \( n = 3 \)
- If \( r = 2 \), then \( n = 6 \)
- If \( r = 3 \), then \( n = 9 \)
- If \( r = 4 \), then \( n = 12 \)
- If \( r = 5 \), then \( n = 15 \)
- Continuing this pattern, we can see that \( n \) can be any multiple of 3.

### Conclusion
Thus, the values of \( n \) can be \( 3, 6, 9, 12, 15, \ldots \), which can be expressed as \( n = 3k \) for any positive integer \( k \).