Let `f(n) = [(1)/(2) + (n)/(100)]` where [n] denotes the integral part of n. Then the value of `sum_(n=1)^(100) f(n)` is

Options

A. 50
B. 51
C. 1
D. 0

Answer

To solve the problem, we need to evaluate the summation of the function \( f(n) = \left\lfloor \frac{1}{2} + \frac{n}{100} \right\rfloor \) from \( n = 1 \) to \( n = 100 \). Here, \( \lfloor x \rfloor \) denotes the integral part of \( x \).

### Step-by-Step Solution:

1. **Understanding the Function**:
The function can be rewritten as:
\[ f(n) = \left\lfloor 0.5 + \frac{n}{100} \right\rfloor \]
This means we need to evaluate \( 0.5 + \frac{n}{100} \) and take the integral part.

2. **Evaluating \( f(n) \) for \( n = 1 \) to \( n = 49 \)**:
- For \( n = 1 \):
\[ f(1) = \left\lfloor 0.5 + \frac{1}{100} \right\rfloor = \left\lfloor 0.5 + 0.01 \right\rfloor = \left\lfloor 0.51 \right\rfloor = 0 \]
- For \( n = 2 \):
\[ f(2) = \left\lfloor 0.5 + \frac{2}{100} \right\rfloor = \left\lfloor 0.5 + 0.02 \right\rfloor = \left\lfloor 0.52 \right\rfloor = 0 \]
- Continuing this way, for \( n = 1 \) to \( n = 49 \), we find:
\[ f(n) = 0 \quad \text{for all } n = 1, 2, \ldots, 49 \]
- Thus, the summation from \( n = 1 \) to \( n = 49 \) is:
\[ \sum_{n=1}^{49} f(n) = 0 \]

3. **Evaluating \( f(n) \) for \( n = 50 \) to \( n = 100 \)**:
- For \( n = 50 \):
\[ f(50) = \left\lfloor 0.5 + \frac{50}{100} \right\rfloor = \left\lfloor 0.5 + 0.5 \right\rfloor = \left\lfloor 1 \right\rfloor = 1 \]
- For \( n = 51 \):
\[ f(51) = \left\lfloor 0.5 + \frac{51}{100} \right\rfloor = \left\lfloor 0.5 + 0.51 \right\rfloor = \left\lfloor 1.01 \right\rfloor = 1 \]
- Continuing this way, for \( n = 50 \) to \( n = 100 \), we find:
\[ f(n) = 1 \quad \text{for all } n = 50, 51, \ldots, 100 \]
- Thus, the summation from \( n = 50 \) to \( n = 100 \) is:
\[ \sum_{n=50}^{100} f(n) = 51 \times 1 = 51 \]

4. **Combining the Results**:
Now, we combine the two parts:
\[ \sum_{n=1}^{100} f(n) = \sum_{n=1}^{49} f(n) + \sum_{n=50}^{100} f(n) = 0 + 51 = 51 \]

### Final Answer:
The value of \( \sum_{n=1}^{100} f(n) \) is \( \boxed{51} \).