If the expansion of `(x-(1)/(x^(2)))^(2n)` contains a term independent of x, then n is a multiple of

Options

A. 2
B. 3
C. 4
D. 5

Answer

To solve the problem of finding the value of \( n \) such that the expansion of \( (x - \frac{1}{x^2})^{2n} \) contains a term independent of \( x \), we will follow these steps:

### Step 1: Identify the General Term
The general term in the binomial expansion of \( (a + b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
In our case, \( a = x \) and \( b = -\frac{1}{x^2} \), and \( n = 2n \). Thus, the general term becomes:
\[ T_{r+1} = \binom{2n}{r} x^{2n - r} \left(-\frac{1}{x^2}\right)^r \]

### Step 2: Simplify the General Term
Now, simplifying the general term:
\[ T_{r+1} = \binom{2n}{r} x^{2n - r} \cdot \left(-1\right)^r \cdot \frac{1}{x^{2r}} = \binom{2n}{r} (-1)^r x^{2n - r - 2r} \]
This simplifies to:
\[ T_{r+1} = \binom{2n}{r} (-1)^r x^{2n - 3r} \]

### Step 3: Find the Condition for the Term to be Independent of \( x \)
For the term to be independent of \( x \), the exponent of \( x \) must be zero:
\[ 2n - 3r = 0 \]
From this equation, we can solve for \( r \):
\[ 2n = 3r \quad \Rightarrow \quad r = \frac{2n}{3} \]

### Step 4: Ensure \( r \) is an Integer
Since \( r \) must be a non-negative integer, \( \frac{2n}{3} \) must also be an integer. This implies that \( 2n \) must be divisible by \( 3 \). Therefore, \( n \) must be a multiple of \( 3 \).

### Conclusion
Thus, the value of \( n \) must be a multiple of \( 3 \).

### Final Answer
The answer is that \( n \) is a multiple of \( 3 \).