For any positive real number R, [R] is the greatest interger less than or equal to R, then the value of `[R/2] + [(R + 1)/(2)]` is :

Options

A. [R]
B. [2R]
C. `[R + 1/3]`
D. can't be determined

Answer

To solve the problem, we need to evaluate the expression \([R/2] + [(R + 1)/2]\) where \([R]\) denotes the greatest integer less than or equal to \(R\).

### Step-by-Step Solution:

1. **Understanding the Floor Function**:
The notation \([R]\) represents the greatest integer less than or equal to \(R\). For example, if \(R = 3.7\), then \([R] = 3\), and if \(R = 4\), then \([R] = 4\).

2. **Expressing \(R\)**:
Let \(n = [R]\). This means \(n\) is the greatest integer less than or equal to \(R\). Therefore, we can write:
\[ n \leq R < n + 1 \]

3. **Calculating \([R/2]\)**:
Since \(R\) is between \(n\) and \(n + 1\), we can find \([R/2]\):
- If \(R\) is in the range \([n, n + 1)\), then:
\[ \frac{n}{2} \leq \frac{R}{2} < \frac{n + 1}{2} \]
- Thus, \([R/2] = \left[\frac{n}{2}\right]\).

4. **Calculating \([(R + 1)/2]\)**:
Now we calculate \([(R + 1)/2]\):
- Since \(R + 1\) is in the range \([n + 1, n + 2)\), we have:
\[ \frac{n + 1}{2} \leq \frac{R + 1}{2} < \frac{n + 2}{2} \]
- Thus, \([(R + 1)/2] = \left[\frac{n + 1}{2}\right]\).

5. **Combining the Results**:
Now we can combine the results:
\[ [R/2] + [(R + 1)/2] = \left[\frac{n}{2}\right] + \left[\frac{n + 1}{2}\right] \]

6. **Analyzing Cases for \(n\)**:
- **Case 1**: If \(n\) is even, say \(n = 2k\):
\[ \left[\frac{2k}{2}\right] + \left[\frac{2k + 1}{2}\right] = k + k = 2k = n \]
- **Case 2**: If \(n\) is odd, say \(n = 2k + 1\):
\[ \left[\frac{2k + 1}{2}\right] + \left[\frac{2k + 2}{2}\right] = k + (k + 1) = 2k + 1 = n \]

7. **Conclusion**:
In both cases, we find that \([R/2] + [(R + 1)/2] = n = [R]\). Therefore, the value of \([R/2] + [(R + 1)/2]\) is equal to \([R]\).

### Final Answer:
\[ [R/2] + [(R + 1)/2] = [R] \]