If f (x) is a function such that f(x+y)=f(x)+f(y) and `f(1)=7" then "underset(r=1)overset(n)sum f(r)` is equal to :

Options

A. 7n/2
B. 7(n+1)2
C. 7n(n+1)
D. 7n(n+1)/2

Answer

To solve the problem, we need to find the value of the summation \( \sum_{r=1}^{n} f(r) \) given the functional equation \( f(x+y) = f(x) + f(y) \) and the condition \( f(1) = 7 \).

### Step-by-Step Solution:

1. **Understanding the Functional Equation**:
The equation \( f(x+y) = f(x) + f(y) \) suggests that \( f \) is a linear function. A common form of such functions is \( f(x) = kx \) for some constant \( k \).

2. **Finding the Constant \( k \)**:
We know that \( f(1) = 7 \). If we assume \( f(x) = kx \), then:
\[ f(1) = k \cdot 1 = k \]
Therefore, \( k = 7 \). This gives us:
\[ f(x) = 7x \]

3. **Calculating \( f(r) \)**:
Now, we can express \( f(r) \):
\[ f(r) = 7r \]

4. **Setting Up the Summation**:
We need to find:
\[ \sum_{r=1}^{n} f(r) = \sum_{r=1}^{n} 7r \]
We can factor out the constant:
\[ = 7 \sum_{r=1}^{n} r \]

5. **Using the Formula for the Sum of First \( n \) Natural Numbers**:
The formula for the sum of the first \( n \) natural numbers is:
\[ \sum_{r=1}^{n} r = \frac{n(n+1)}{2} \]
Therefore:
\[ \sum_{r=1}^{n} f(r) = 7 \cdot \frac{n(n+1)}{2} \]

6. **Final Result**:
Thus, the final result is:
\[ \sum_{r=1}^{n} f(r) = \frac{7n(n+1)}{2} \]

### Conclusion:
The value of \( \sum_{r=1}^{n} f(r) \) is \( \frac{7n(n+1)}{2} \).