If f(x) is a differntiable function satisfying the condition `f(100x)=x+f(100x-100)`, `forall x in R` and `f(100)=1`, then `f(10^(4))` is

Options

A. 5049
B. `sum_(r=1)^(100)r`
C. `sum_(r=2)^(100)r`
D. 5050

Answer

To solve the problem, we need to find the value of \( f(10^4) \) given the functional equation \( f(100x) = x + f(100x - 100) \) and the initial condition \( f(100) = 1 \).

### Step-by-Step Solution:

1. **Substituting \( x = 1 \)**:
\[ f(100 \cdot 1) = 1 + f(100 \cdot 1 - 100) \]
This simplifies to:
\[ f(100) = 1 + f(0) \]
Since \( f(100) = 1 \), we have:
\[ 1 = 1 + f(0) \implies f(0) = 0 \]

2. **Substituting \( x = 2 \)**:
\[ f(100 \cdot 2) = 2 + f(100 \cdot 2 - 100) \]
This simplifies to:
\[ f(200) = 2 + f(100) \]
Since \( f(100) = 1 \), we have:
\[ f(200) = 2 + 1 = 3 \]

3. **Substituting \( x = 3 \)**:
\[ f(100 \cdot 3) = 3 + f(100 \cdot 3 - 100) \]
This simplifies to:
\[ f(300) = 3 + f(200) \]
Since \( f(200) = 3 \), we have:
\[ f(300) = 3 + 3 = 6 \]

4. **Substituting \( x = 4 \)**:
\[ f(100 \cdot 4) = 4 + f(100 \cdot 4 - 100) \]
This simplifies to:
\[ f(400) = 4 + f(300) \]
Since \( f(300) = 6 \), we have:
\[ f(400) = 4 + 6 = 10 \]

5. **Identifying the pattern**:
We observe the values:
- \( f(100) = 1 \)
- \( f(200) = 3 \)
- \( f(300) = 6 \)
- \( f(400) = 10 \)

The values \( 1, 3, 6, 10 \) correspond to the triangular numbers, specifically:
\[ f(100n) = \frac{n(n+1)}{2} \]
where \( n \) is the integer multiplier of 100.

6. **Finding \( f(10^4) \)**:
Since \( 10^4 = 100 \cdot 100 \), we set \( n = 100 \):
\[ f(10^4) = f(100 \cdot 100) = \frac{100(100 + 1)}{2} = \frac{100 \cdot 101}{2} = 5050 \]

### Final Answer:
Thus, the value of \( f(10^4) \) is \( 5050 \).