Let f be a function satisfying `f(x+y)=f(x) + f(y)` for all `x,y in R`. If `f (1)= k` then `f(n), n in N` is equal to

Options

A. `k^(n)`
B. nk
C. k
D. None of these

Answer

To solve the problem, we need to find a general expression for \( f(n) \) where \( n \) is a natural number, given that \( f \) satisfies the functional equation \( f(x+y) = f(x) + f(y) \) for all \( x, y \in \mathbb{R} \) and \( f(1) = k \).

### Step-by-Step Solution:

1. **Understanding the Functional Equation**:
The equation \( f(x+y) = f(x) + f(y) \) indicates that \( f \) is a linear function. This property is known as Cauchy's functional equation.

2. **Substituting Values**:
Let's substitute \( x = 1 \) and \( y = 1 \) into the functional equation:
\[ f(1 + 1) = f(1) + f(1) \]
This simplifies to:
\[ f(2) = f(1) + f(1) = 2f(1) \]
Since \( f(1) = k \), we have:
\[ f(2) = 2k \]

3. **Finding \( f(3) \)**:
Now, let's find \( f(3) \) by substituting \( x = 2 \) and \( y = 1 \):
\[ f(2 + 1) = f(2) + f(1) \]
This gives us:
\[ f(3) = f(2) + f(1) = 2k + k = 3k \]

4. **Continuing the Pattern**:
We can continue this process:
- For \( f(4) \):
\[ f(3 + 1) = f(3) + f(1) \implies f(4) = 3k + k = 4k \]
- For \( f(5) \):
\[ f(4 + 1) = f(4) + f(1) \implies f(5) = 4k + k = 5k \]

5. **Generalizing the Result**:
From the pattern observed, we can generalize that:
\[ f(n) = nk \quad \text{for any natural number } n. \]

### Final Result:
Thus, we conclude that:
\[ f(n) = nk \quad \text{for all } n \in \mathbb{N}. \]