Let f be a real valued function satisfying `f(x+y)=f(x)+f(y)` for all ` x, y in R and f(1)=2`. Then `sum_(k=1)^(n)f(k)=`

Options

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

Explanation

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 \( x, y \in \mathbb{R} \) and \( f(1) = 2 \). This type of functional equation suggests that \( f \) is a linear function.

### Step 2: Find values of \( f \) for integers
Let's start by finding \( f(2) \):
- Set \( x = 1 \) and \( y = 1 \):
\[ f(2) = f(1 + 1) = f(1) + f(1) = 2 + 2 = 4 \]

Next, find \( f(3) \):
- Set \( x = 2 \) and \( y = 1 \):
\[ f(3) = f(2 + 1) = f(2) + f(1) = 4 + 2 = 6 \]

Now, find \( f(4) \):
- Set \( x = 2 \) and \( y = 2 \):
\[ f(4) = f(2 + 2) = f(2) + f(2) = 4 + 4 = 8 \]

### Step 3: Identify a pattern
From our calculations:
- \( f(1) = 2 \)
- \( f(2) = 4 \)
- \( f(3) = 6 \)
- \( f(4) = 8 \)

We can see a pattern forming:
\[ f(n) = 2n \quad \text{for } n = 1, 2, 3, 4, \ldots \]

### Step 4: Generalize the function
We can hypothesize that \( f(n) = 2n \) for all integers \( n \). We can prove this by induction or by recognizing that the functional equation \( f(x+y) = f(x) + f(y) \) implies linearity.

### Step 5: Calculate the summation
We need to find:
\[ \sum_{k=1}^{n} f(k) = \sum_{k=1}^{n} 2k \]
This simplifies to:
\[ 2 \sum_{k=1}^{n} k \]
We know the formula for the sum of the first \( n \) natural numbers:
\[ \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \]
Thus,
\[ \sum_{k=1}^{n} f(k) = 2 \cdot \frac{n(n+1)}{2} = n(n+1) \]

### Conclusion
The final answer is:
\[ \sum_{k=1}^{n} f(k) = n(n+1) \]