If `f(n) = [(1)/(3)+(n)/(100)]`, where `[.]` denotes `G.I.F` then `sum_(n=1)^(200)f(n)` is equal to

Options

A. `184`
B. `165`
C. `167`
D. `168`

Explanation

To solve the problem, we need to evaluate the summation \( \sum_{n=1}^{200} f(n) \) where \( f(n) = \left\lfloor \frac{1}{3} + \frac{n}{100} \right\rfloor \).

### Step-by-Step Solution:

1. **Understanding the Function**:
The function \( f(n) \) is defined as:
\[ f(n) = \left\lfloor \frac{1}{3} + \frac{n}{100} \right\rfloor \]
The term \( \frac{1}{3} \) is approximately \( 0.333 \).

2. **Finding the Range for \( n \)**:
We need to calculate \( f(n) \) for \( n = 1, 2, \ldots, 200 \). First, we can express \( f(n) \) as:
\[ f(n) = \left\lfloor 0.333 + \frac{n}{100} \right\rfloor \]
This means we need to find the integer part of \( 0.333 + \frac{n}{100} \).

3. **Calculating Critical Points**:
To find when \( f(n) \) changes its value, we set:
\[ 0.333 + \frac{n}{100} = k \quad \text{(where \( k \) is an integer)} \]
Rearranging gives:
\[ n = 100(k - 0.333) \]
This means \( n \) will change its integer value at \( n = 100k - 33.3 \).

4. **Finding Values of \( k \)**:
We calculate \( k \) for different ranges of \( n \):
- For \( n = 1 \) to \( n = 66 \):
\[ f(n) = 0 \quad \text{(since \( 0.333 + \frac{n}{100} < 1 \))} \]
- For \( n = 67 \) to \( n = 166 \):
\[ f(n) = 1 \quad \text{(since \( 1 \leq 0.333 + \frac{n}{100} < 2 \))} \]
- For \( n = 167 \) to \( n = 200 \):
\[ f(n) = 2 \quad \text{(since \( 2 \leq 0.333 + \frac{n}{100} < 3 \))} \]

5. **Calculating the Summation**:
Now, we calculate the contributions to the sum:
- From \( n = 1 \) to \( n = 66 \):
\[ \text{Contribution} = 66 \times 0 = 0 \]
- From \( n = 67 \) to \( n = 166 \):
\[ \text{Contribution} = 100 \times 1 = 100 \]
- From \( n = 167 \) to \( n = 200 \):
\[ \text{Contribution} = 34 \times 2 = 68 \]

6. **Final Sum**:
Adding all contributions together:
\[ \text{Total Sum} = 0 + 100 + 68 = 168 \]

### Conclusion:
Thus, the value of \( \sum_{n=1}^{200} f(n) \) is \( \boxed{168} \).