Let `f(x)` be a function such that `: f(x - 1) + f(x + 1) = sqrt3 f(x),` for all `x epsilon R.` If `f(5) = 100,` then prove that the value of `sum_(r=o)^99 f(5 + 12r)` will be equal to `10000.`

Answer

To solve the problem, we need to analyze the functional equation given and use it to find the sum \( \sum_{r=0}^{99} f(5 + 12r) \).

### Step-by-Step Solution:

1. **Understand the Functional Equation**:
The functional equation given is:
\[ f(x - 1) + f(x + 1) = \sqrt{3} f(x) \]
This equation holds for all \( x \in \mathbb{R} \).

2. **Find the Periodicity of the Function**:
To explore the periodicity, we can substitute \( x \) with \( x + 1 \) and \( x - 1 \) in the functional equation:
- For \( x + 1 \):
\[ f(x) + f(x + 2) = \sqrt{3} f(x + 1) \]
- For \( x - 1 \):
\[ f(x - 2) + f(x) = \sqrt{3} f(x - 1) \]

By manipulating these equations, we can derive that:
\[ f(x + 2) + f(x - 2) = \sqrt{3} f(x) \]
Continuing this process, we can show that:
\[ f(x + 12) = f(x) \]
Thus, the function \( f(x) \) is periodic with a period of 12.

3. **Evaluate the Sum**:
We need to evaluate:
\[ \sum_{r=0}^{99} f(5 + 12r) \]
Since \( f(x) \) is periodic with period 12, we have:
\[ f(5 + 12r) = f(5) \]
for all integers \( r \). This means:
\[ f(5 + 12r) = f(5) = 100 \]

4. **Compute the Total**:
The sum can now be simplified:
\[ \sum_{r=0}^{99} f(5 + 12r) = \sum_{r=0}^{99} 100 = 100 \times 100 = 10000 \]

5. **Conclusion**:
Therefore, we have shown that:
\[ \sum_{r=0}^{99} f(5 + 12r) = 10000 \]