If `f(x)=1-x+x^(2)-x^(3)+ . . .-x^(99)+x^(100)` then `f'(1)` is equal to

Options

A. A. 100
B. B. 50
C. C. 49
D. D. 51

Answer

To solve the problem, we need to find the derivative of the function \( f(x) = 1 - x + x^2 - x^3 + \ldots - x^{99} + x^{100} \) and then evaluate \( f'(1) \).

### Step-by-Step Solution:

1. **Identify the function**:
The function is given as:
\[ f(x) = 1 - x + x^2 - x^3 + \ldots - x^{99} + x^{100} \]

2. **Recognize the series**:
This is a finite series where the signs alternate. We can express it as:
\[ f(x) = \sum_{n=0}^{100} (-1)^n x^n \]

3. **Differentiate the function**:
To find \( f'(x) \), we differentiate term by term:
\[ f'(x) = 0 - 1 + 2x - 3x^2 + 4x^3 - \ldots - 99x^{98} + 100x^{99} \]
This can be simplified as:
\[ f'(x) = -1 + 2x - 3x^2 + 4x^3 - \ldots - 99x^{98} + 100x^{99} \]

4. **Evaluate \( f'(1) \)**:
Now, we substitute \( x = 1 \) into the derivative:
\[ f'(1) = -1 + 2(1) - 3(1^2) + 4(1^3) - \ldots - 99(1^{98}) + 100(1^{99}) \]
This simplifies to:
\[ f'(1) = -1 + 2 - 3 + 4 - 5 + \ldots - 99 + 100 \]

5. **Group the terms**:
We can group the terms into pairs:
\[ (-1 + 2) + (-3 + 4) + \ldots + (-99 + 100) \]
Each pair sums to 1, and since there are 100 terms, we have 50 pairs:
\[ \text{Number of pairs} = \frac{100}{2} = 50 \]
Therefore, the total sum is:
\[ f'(1) = 50 \]

### Final Answer:
Thus, \( f'(1) = 50 \).