If `f(x)=1-x+x^2-x^3+......-x^(99)+x^(100)` then `f^(prime)(1)` equals

Options

A. 150
B. `-50`
C. `-150`
D. 50

Explanation

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 it at \( x = 1 \).

### 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. **Differentiate the Function**:
To find \( f'(x) \), we differentiate each term of the function:
- The derivative of \( 1 \) is \( 0 \).
- The derivative of \( -x \) is \( -1 \).
- The derivative of \( x^2 \) is \( 2x \).
- The derivative of \( -x^3 \) is \( -3x^2 \).
- Continuing this way, the derivative of \( x^n \) is \( nx^{n-1} \).

Therefore, the derivative \( f'(x) \) becomes:
\[ f'(x) = 0 - 1 + 2x - 3x^2 + 4x^3 - \ldots - 99x^{98} + 100x^{99} \]

3. **Reorganize the Derivative**:
We can rearrange the terms in \( f'(x) \):
\[ f'(x) = -1 + 2x - 3x^2 + 4x^3 - \ldots - 99x^{98} + 100x^{99} \]

4. **Evaluate the Derivative at \( x = 1 \)**:
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. **Separate Odd and Even Terms**:
We can group the terms into odd and even:
\[ f'(1) = (2 + 4 + 6 + \ldots + 100) + (-1 - 3 - 5 - \ldots - 99) \]

6. **Calculate the Sum of Even Terms**:
The sum of the first \( n \) even numbers is given by:
\[ S_{even} = n(n + 1) \]
where \( n \) is the number of even terms. Here, there are 50 even terms (from 2 to 100):
\[ S_{even} = 50 \times 51 = 2550 \]

7. **Calculate the Sum of Odd Terms**:
The sum of the first \( n \) odd numbers is given by:
\[ S_{odd} = n^2 \]
where \( n \) is the number of odd terms. Here, there are 50 odd terms (from 1 to 99):
\[ S_{odd} = 50^2 = 2500 \]

8. **Combine the Sums**:
Now, substituting back into our equation:
\[ f'(1) = 2550 - 2500 = 50 \]

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