If `x^(2k)` occurs in the expansion of `(x+1/(x^(2)))^(n-3)`, then

Options

A. n - 2k is a multiple of 2
B. n- 2k is a multtiple of 3
C. k=0
D. none of these

Answer

To solve the problem, we need to find the conditions under which \( x^{2k} \) occurs in the expansion of \( \left( x + \frac{1}{x^2} \right)^{n-3} \).

### Step-by-Step Solution:

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

2. **Simplify the General Term**:
Simplifying the term:
\[ T_{r+1} = \binom{n-3}{r} x^{n-3-r} \cdot x^{-2r} = \binom{n-3}{r} x^{n-3-3r} \]

3. **Set the Power of x**:
We want the term \( x^{2k} \) to exist in the expansion. Therefore, we set:
\[ n - 3 - 3r = 2k \]

4. **Rearranging the Equation**:
Rearranging gives us:
\[ n - 2k = 3r + 3 \]
or
\[ n - 2k - 3 = 3r \]

5. **Determine Conditions**:
From the equation \( n - 2k - 3 = 3r \), we can see that \( n - 2k - 3 \) must be a multiple of 3. This implies:
\[ n - 2k \equiv 0 \mod 3 \]

6. **Conclusion**:
Therefore, the condition that must hold true for \( x^{2k} \) to occur in the expansion is that \( n - 2k \) is a multiple of 3.

### Final Answer:
The correct condition is:
- \( n - 2k \) is a multiple of 3.