`If `
`f(x+ y) = f (x)+ f (y) ` for all `x,y in R` and f (1) = ` lamda `, then `f(n)`, `n in N` is

Options

A. `lamda^n`
B. ` (1)/(lamda^n)`
C. `n lamda`
D. None

Answer

To solve the problem, we start with the functional equation given:

\[ f(x + y) = f(x) + f(y) \]

for all \( x, y \in \mathbb{R} \), and we know that \( f(1) = \lambda \).

### Step 1: Find \( f(2) \)

Let's set \( x = 1 \) and \( y = 1 \) in the functional equation:

\[ f(1 + 1) = f(1) + f(1) \]

This simplifies to:

\[ f(2) = f(1) + f(1) = 2f(1) = 2\lambda \]

### Step 2: Find \( f(3) \)

Now, let's set \( x = 2 \) and \( y = 1 \):

\[ f(2 + 1) = f(2) + f(1) \]

This simplifies to:

\[ f(3) = f(2) + f(1) = 2\lambda + \lambda = 3\lambda \]

### Step 3: Find \( f(4) \)

Next, we set \( x = 3 \) and \( y = 1 \):

\[ f(3 + 1) = f(3) + f(1) \]

This simplifies to:

\[ f(4) = f(3) + f(1) = 3\lambda + \lambda = 4\lambda \]

### Step 4: Generalize for \( f(n) \)

From the pattern we observe:

- \( f(1) = \lambda \)
- \( f(2) = 2\lambda \)
- \( f(3) = 3\lambda \)
- \( f(4) = 4\lambda \)

We can see that for any natural number \( n \):

\[ f(n) = n\lambda \]

### Conclusion

Thus, the value of \( f(n) \) for \( n \in \mathbb{N} \) is:

\[ f(n) = n\lambda \]