If `f(x)=(x-1)^(100)(x-2)^(2(99))(x-3)^(3(98))…(x-100)^(100),` then the value of `(f'(101))/(f(101))` is

Options

A. 5050
B. 2575
C. 3030
D. 1250

Explanation

To solve the problem, we need to find the value of \(\frac{f'(101)}{f(101)}\) where

\[ f(x) = (x-1)^{100} (x-2)^{2(99)} (x-3)^{3(98)} \cdots (x-100)^{100}. \]

### Step 1: Rewrite the function in a summation form

We can express \(f(x)\) in a more general form:

\[ f(x) = \prod_{i=1}^{100} (x-i)^{i(101-i)}. \]

### Step 2: Take the logarithm of \(f(x)\)

Taking the logarithm of both sides gives:

\[ \log f(x) = \sum_{i=1}^{100} i(101-i) \log(x-i). \]

### Step 3: Differentiate both sides with respect to \(x\)

Using the chain rule, we differentiate:

\[ \frac{f'(x)}{f(x)} = \sum_{i=1}^{100} i(101-i) \cdot \frac{1}{x-i}. \]

### Step 4: Substitute \(x = 101\)

Now we need to evaluate this at \(x = 101\):

\[ \frac{f'(101)}{f(101)} = \sum_{i=1}^{100} i(101-i) \cdot \frac{1}{101-i}. \]

### Step 5: Simplify the summation

Notice that \( \frac{1}{101-i} \) cancels out with \( (101-i) \):

\[ \frac{f'(101)}{f(101)} = \sum_{i=1}^{100} i. \]

### Step 6: Calculate the sum of the first 100 natural numbers

The sum of the first \(n\) natural numbers is given by the formula:

\[ \sum_{i=1}^{n} i = \frac{n(n+1)}{2}. \]

For \(n = 100\):

\[ \sum_{i=1}^{100} i = \frac{100 \cdot 101}{2} = 5050. \]

### Final Answer

Thus, the value of \(\frac{f'(101)}{f(101)}\) is

\[ \boxed{5050}. \]