`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. `2^(n)`
C. `n2^(n-1)`
D. `-n2^(n - 1)`

Explanation

To solve the expression \( 1^2 C_1 - 2^2 C_2 + 3^2 C_3 - 4^2 C_4 + \ldots + (-1)^{n-2} n^2 C_n \), we can use the Binomial Theorem and properties of binomial coefficients.

### Step-by-Step Solution:

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

2. **Substituting \( x = -1 \)**:
If we substitute \( x = -1 \) into the binomial expansion, we have:
\[ (1 - 1)^n = 0 \]
This gives us:
\[ C_0 - C_1 + C_2 - C_3 + \ldots + (-1)^n C_n = 0 \]

3. **Rearranging the Terms**:
Rearranging the above equation, we can express it as:
\[ C_1 - C_2 + C_3 - C_4 + \ldots + (-1)^{n-1} C_n = C_0 \]

4. **Differentiating the Binomial Expansion**:
Now, we differentiate the binomial expansion with respect to \( x \):
\[ \frac{d}{dx}[(1 + x)^n] = n(1 + x)^{n-1} \]
The derivative of the right-hand side gives:
\[ C_1 + 2C_2 x + 3C_3 x^2 + \ldots + nC_n x^{n-1} \]

5. **Substituting \( x = -1 \) Again**:
Now substituting \( x = -1 \) into the differentiated equation:
\[ n(1 - 1)^{n-1} = 0 \]
This gives:
\[ C_1 - 2C_2 + 3C_3 - 4C_4 + \ldots + (-1)^{n-1} n C_n = 0 \]

6. **Multiplying by \( x \)**:
To find the expression \( 1^2 C_1 - 2^2 C_2 + 3^2 C_3 - 4^2 C_4 + \ldots \), we can multiply the previous result by \( x \) and differentiate again:
\[ \frac{d^2}{dx^2}[(1 + x)^n] = n(n-1)(1 + x)^{n-2} \]
The second derivative gives:
\[ C_1 + 4C_2 x + 9C_3 x^2 + \ldots + n^2 C_n x^{n-1} \]

7. **Final Evaluation**:
Substituting \( x = -1 \) again, we find:
\[ n(n-1)(1 - 1)^{n-2} = 0 \]
Thus, the expression simplifies to:
\[ 1^2 C_1 - 2^2 C_2 + 3^2 C_3 - 4^2 C_4 + \ldots + (-1)^{n-2} n^2 C_n = 0 \]

### Conclusion:
The final result of the expression is:
\[ \boxed{0} \]