If `f(x)=x^(100)+x^(99)++x+1,` then `f^(prime)(1)` is equal to

Options

A. 5050
B. 5049
C. 5051
D. 50051

Explanation

To find \( f'(1) \) for the function \( f(x) = x^{100} + x^{99} + x^{98} + \ldots + x + 1 \), we will follow these steps:

### Step 1: Differentiate the function \( f(x) \)

The function can be expressed as:
\[ f(x) = x^{100} + x^{99} + x^{98} + \ldots + x + 1 \]

To differentiate \( f(x) \), we apply the power rule of differentiation:
\[ f'(x) = 100x^{99} + 99x^{98} + 98x^{97} + \ldots + 2x + 1 \]

### Step 2: Evaluate \( f'(1) \)

Now we need to find \( f'(1) \):
\[ f'(1) = 100 \cdot 1^{99} + 99 \cdot 1^{98} + 98 \cdot 1^{97} + \ldots + 2 \cdot 1 + 1 \]

Since \( 1^n = 1 \) for any \( n \), we can simplify this to:
\[ f'(1) = 100 + 99 + 98 + \ldots + 2 + 1 \]

### Step 3: Calculate the sum of the series

The sum \( 100 + 99 + 98 + \ldots + 2 + 1 \) is the sum of the first 100 natural numbers. The formula for the sum of the first \( n \) natural numbers is:
\[ S_n = \frac{n(n + 1)}{2} \]
where \( n = 100 \).

Substituting \( n = 100 \) into the formula gives:
\[ S_{100} = \frac{100 \cdot (100 + 1)}{2} = \frac{100 \cdot 101}{2} = 50 \cdot 101 = 5050 \]

### Final Result

Thus, we find that:
\[ f'(1) = 5050 \]