If `f (x+1)+ f (x-1)=2 f (x) and f(0)=0,` then f` (n),n in N `is

Options

A. nf(1)
B. `[ f (1)]^n`
C. ` 0`
D. None

Answer

To solve the problem, we will follow a systematic approach based on the functional equation given.

### Step-by-Step Solution:

1. **Understanding the Functional Equation**:
We have the equation:
\[ f(x+1) + f(x-1) = 2f(x) \]
This equation suggests that the function \( f(x) \) is linear.

2. **Finding Values of f**:
We are given that \( f(0) = 0 \). We will use this information to find values of \( f \) for other integers.

3. **Substituting Values**:
- **For \( x = 1 \)**:
\[ f(2) + f(0) = 2f(1) \implies f(2) + 0 = 2f(1) \implies f(2) = 2f(1) \]

- **For \( x = 2 \)**:
\[ f(3) + f(1) = 2f(2) \implies f(3) + f(1) = 2(2f(1)) \implies f(3) + f(1) = 4f(1) \]
Rearranging gives:
\[ f(3) = 4f(1) - f(1) = 3f(1) \]

- **For \( x = 3 \)**:
\[ f(4) + f(2) = 2f(3) \implies f(4) + 2f(1) = 2(3f(1)) \implies f(4) + 2f(1) = 6f(1) \]
Rearranging gives:
\[ f(4) = 6f(1) - 2f(1) = 4f(1) \]

4. **Identifying the Pattern**:
From the calculations:
- \( f(1) = f(1) \)
- \( f(2) = 2f(1) \)
- \( f(3) = 3f(1) \)
- \( f(4) = 4f(1) \)

We can see a pattern emerging:
\[ f(n) = n f(1) \quad \text{for } n \in \mathbb{N} \]

5. **Conclusion**:
Therefore, we can generalize that:
\[ f(n) = n f(1) \]
where \( f(1) \) is a constant.

### Final Answer:
Thus, \( f(n) \) for \( n \in \mathbb{N} \) is:
\[ f(n) = n \cdot f(1) \]