The middle term in the expansion of `(1-3x+3x^2-x^3)^(2n)` is (a) `((6n)!x^(n))/((3n)!(3n)!)` (b) `((6n)!x^(3n))/((3n)!)` (c) `((6n)!)/((3n)!(3n)!)(-x)^(3n)` (d) none of these

Options

A. `((6n)!x^(n))/((3n)!(3n)!)`
B. `((6n)!x^(3n))/((3n)!)`
C. `((6n))/((3n)!(3n)!)(-x)^(3n)`
D. None of these

Explanation

To find the middle term in the expansion of \((1 - 3x + 3x^2 - x^3)^{2n}\), we can follow these steps:

### Step 1: Recognize the expression as a binomial
The expression \(1 - 3x + 3x^2 - x^3\) can be rewritten as a cube. We can express it as:
\[ (1 - x)^3 \]
This is because:
\[ (1 - x)^3 = 1 - 3x + 3x^2 - x^3 \]

### Step 2: Rewrite the original expression
Thus, we can rewrite the original expression:
\[ (1 - 3x + 3x^2 - x^3)^{2n} = ((1 - x)^3)^{2n} = (1 - x)^{6n} \]

### Step 3: Identify the middle term
In the expansion of \((1 - x)^{6n}\), the total number of terms is \(6n + 1\) (since the power is \(6n\)). Since \(6n\) is even, the middle term will be the \((3n + 1)\)-th term.

### Step 4: Use the binomial theorem
According to the binomial theorem, the \(r\)-th term in the expansion of \((a + b)^n\) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
In our case, \(a = 1\), \(b = -x\), and \(n = 6n\). We need to find the \(3n + 1\)-th term, which corresponds to \(r = 3n\).

### Step 5: Calculate the middle term
Thus, the middle term is:
\[ T_{3n + 1} = \binom{6n}{3n} (1)^{6n - 3n} (-x)^{3n} = \binom{6n}{3n} (-x)^{3n} \]

### Step 6: Write the final expression
The middle term can be expressed as:
\[ \frac{(6n)!}{(3n)!(3n)!} (-x)^{3n} \]

### Conclusion
Thus, the middle term in the expansion of \((1 - 3x + 3x^2 - x^3)^{2n}\) is:
\[ \frac{(6n)!}{(3n)!(3n)!} (-x)^{3n} \]

### Step 7: Identify the correct option
Comparing with the given options, we see that this corresponds to option (c):
\[ \frac{(6n)!}{(3n)!(3n)!} (-x)^{3n} \]