If `f : R rarr R` satisfies `f(x + y) = f(x) + f(y) x, y in R` and f(1) = 7, then `sum_(r = 1)^(n) f(r) =`

Options

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

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the functional equation
We are given that \( f(x + y) = f(x) + f(y) \) for all \( x, y \in \mathbb{R} \). This property suggests that \( f \) is a linear function.

### Step 2: Find \( f(2) \)
Using the property of the function, we can find \( f(2) \):
\[ f(2) = f(1 + 1) = f(1) + f(1) = 2f(1) = 2 \times 7 = 14. \]

### Step 3: Find \( f(3) \)
Next, we can find \( f(3) \):
\[ f(3) = f(1 + 2) = f(1) + f(2) = 7 + 14 = 21. \]

### Step 4: Generalize \( f(n) \)
Continuing this pattern, we can see that:
\[ f(1) = 7, \quad f(2) = 14, \quad f(3) = 21. \]
We can generalize this to:
\[ f(n) = n \cdot 7. \]

### Step 5: Calculate the summation \( \sum_{r=1}^{n} f(r) \)
Now we need to calculate:
\[ \sum_{r=1}^{n} f(r) = f(1) + f(2) + f(3) + \ldots + f(n). \]
Substituting the expression for \( f(r) \):
\[ \sum_{r=1}^{n} f(r) = 7 + 14 + 21 + \ldots + 7n. \]
Factoring out the 7:
\[ = 7(1 + 2 + 3 + \ldots + n). \]

### Step 6: Use the formula for the sum of the first \( n \) natural numbers
We know that the sum of the first \( n \) natural numbers is given by:
\[ 1 + 2 + 3 + \ldots + n = \frac{n(n + 1)}{2}. \]
Thus, we have:
\[ \sum_{r=1}^{n} f(r) = 7 \cdot \frac{n(n + 1)}{2}. \]

### Final Answer
The final result is:
\[ \sum_{r=1}^{n} f(r) = \frac{7n(n + 1)}{2}. \]