If `C_(0), C_(1), C_(2),..., C_(n)` denote the binomial
coefficients in the expansion of `(1 + x)^(n)` , then . `1^(2). C_(1) - 2^(2) . C_(2)+ 3^(2). C_(3) -4^(2)C_(4) + ...+ (-1).""^(n-2)n^(2)C_(n)=`.

Options

A. 0
B. `n^(2) 2^(n)`
C. `2^(n)`
D. `n(n-1) 2^(n-2)`

Explanation

To solve the problem, we need to evaluate the expression:

\[ 1^2 \cdot C_1 - 2^2 \cdot C_2 + 3^2 \cdot C_3 - 4^2 \cdot C_4 + \ldots + (-1)^{n-2} n^2 \cdot C_n \]

where \( C_k \) are the binomial coefficients from the expansion of \( (1 + x)^n \).

### Step-by-Step Solution:

1. **Understand the Binomial Expansion**:
The binomial expansion of \( (1 + x)^n \) is given by:
\[ (1 + x)^n = C_0 + C_1 x + C_2 x^2 + C_3 x^3 + \ldots + C_n x^n \]
where \( C_k = \binom{n}{k} \).

2. **Set Up the Expression**:
We need to evaluate the expression:
\[ S = 1^2 C_1 - 2^2 C_2 + 3^2 C_3 - 4^2 C_4 + \ldots + (-1)^{n-2} n^2 C_n \]

3. **Use Values of \( x \)**:
To simplify the evaluation, we can use the values \( x = 1 \) and \( x = -1 \) in the binomial expansion.

- For \( x = 1 \):
\[ (1 + 1)^n = 2^n = C_0 + C_1 + C_2 + C_3 + \ldots + C_n \]

- For \( x = -1 \):
\[ (1 - 1)^n = 0 = C_0 - C_1 + C_2 - C_3 + \ldots + (-1)^n C_n \]

4. **Combine the Results**:
The two expansions give us:
\[ C_0 + C_1 + C_2 + C_3 + \ldots + C_n = 2^n \]
\[ C_0 - C_1 + C_2 - C_3 + \ldots + (-1)^n C_n = 0 \]

5. **Multiply by \( x \)**:
Now, differentiate \( (1 + x)^n \) and then multiply by \( x \):
\[ \frac{d}{dx} (1 + x)^n = n(1 + x)^{n-1} \]
Multiplying both sides by \( x \):
\[ x \frac{d}{dx} (1 + x)^n = n x (1 + x)^{n-1} \]

6. **Evaluate at \( x = 1 \)**:
\[ S = n \cdot 1 \cdot (1 + 1)^{n-1} = n \cdot 2^{n-1} \]

7. **Final Step**:
The original expression can be evaluated as:
\[ S = 0 \]
because the alternating sum of the binomial coefficients results in zero when evaluated at \( x = -1 \).

### Conclusion:
Thus, the value of the expression is:
\[ \boxed{0} \]