If `f(x)=1+x+(x^2)/2++(x^(100))/(100),` then `f^(prime)(1)` is equal to

Options

A. `1/(100)`
B. 100
C. 0
D. Does not exist

Explanation

To solve the problem, we need to find the derivative of the function \( f(x) \) and then evaluate it at \( x = 1 \).

### Step-by-Step Solution:

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

2. **Differentiate the Function:**
We will differentiate \( f(x) \) term by term. The derivative of \( x^n \) is \( n \cdot x^{n-1} \). Thus, the derivative \( f'(x) \) is:
\[ f'(x) = 0 + 1 + \frac{2x}{2} + \frac{3x^2}{3} + \ldots + \frac{100x^{99}}{100} \]
Simplifying this, we get:
\[ f'(x) = 1 + 1 + x + x^2 + \ldots + x^{99} \]

3. **Recognize the Pattern:**
The expression \( 1 + 1 + x + x^2 + \ldots + x^{99} \) can be simplified further. The first two terms give us \( 2 \), and the remaining terms form a geometric series:
\[ f'(x) = 2 + x + x^2 + \ldots + x^{99} \]

4. **Sum of the Geometric Series:**
The sum of the geometric series \( x + x^2 + \ldots + x^{99} \) can be calculated using the formula for the sum of a geometric series:
\[ S_n = a \frac{1 - r^n}{1 - r} \]
Here, \( a = x \), \( r = x \), and \( n = 99 \):
\[ S = x \frac{1 - x^{99}}{1 - x} \]
Therefore:
\[ f'(x) = 2 + \frac{x(1 - x^{99})}{1 - x} \]

5. **Evaluate the Derivative at \( x = 1 \):**
Now we need to evaluate \( f'(1) \):
\[ f'(1) = 2 + \frac{1(1 - 1^{99})}{1 - 1} \]
The term \( \frac{1(1 - 1)}{1 - 1} \) is indeterminate, but we can evaluate the limit as \( x \) approaches \( 1 \):
\[ S = 1 + 1 + 1 + \ldots + 1 \quad (\text{100 terms}) \]
Thus:
\[ f'(1) = 100 \]

### Final Answer:
\[ f'(1) = 100 \]