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

Options

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

Explanation

To solve the problem, we need to find the function \( f(x) \) that satisfies the functional equation \( f(x+y) = f(x)f(y) \) for all \( x, y \in \mathbb{R} \) and the condition \( f(1) = 2 \). Then, we will compute the sum \( \sum_{k=1}^{n} f(k) \).

### Step-by-Step Solution:

1. **Understanding the Functional Equation**:
The functional equation \( f(x+y) = f(x)f(y) \) suggests that \( f \) could be an exponential function. A common form for such functions is \( f(x) = a^x \) for some constant \( a \).

2. **Finding \( f(0) \)**:
Let's evaluate \( f(0) \) by setting \( x = 0 \) and \( y = 0 \):
\[ f(0 + 0) = f(0)f(0) \implies f(0) = f(0)^2. \]
This implies \( f(0)(f(0) - 1) = 0 \). Thus, \( f(0) = 0 \) or \( f(0) = 1 \). Since \( f(1) = 2 \) and we want \( f \) to be non-zero, we take \( f(0) = 1 \).

3. **Finding \( f(2) \)**:
Now, we can find \( f(2) \) using the functional equation:
\[ f(2) = f(1 + 1) = f(1)f(1) = 2 \cdot 2 = 4. \]

4. **Finding \( f(3) \)**:
Next, we find \( f(3) \):
\[ f(3) = f(2 + 1) = f(2)f(1) = 4 \cdot 2 = 8. \]

5. **Finding \( f(4) \)**:
Continuing this process, we find \( f(4) \):
\[ f(4) = f(3 + 1) = f(3)f(1) = 8 \cdot 2 = 16. \]

6. **Identifying the Pattern**:
From our calculations, we observe:
- \( f(1) = 2 = 2^1 \)
- \( f(2) = 4 = 2^2 \)
- \( f(3) = 8 = 2^3 \)
- \( f(4) = 16 = 2^4 \)

Thus, we can generalize that \( f(n) = 2^n \) for \( n \in \mathbb{N} \).

7. **Summation**:
Now, we need to compute the sum:
\[ \sum_{k=1}^{n} f(k) = \sum_{k=1}^{n} 2^k. \]
This is a geometric series with first term \( a = 2 \) and common ratio \( r = 2 \). The formula for the sum of the first \( n \) terms of a geometric series is:
\[ S_n = a \frac{r^n - 1}{r - 1}. \]
Applying this:
\[ S_n = 2 \frac{2^n - 1}{2 - 1} = 2(2^n - 1) = 2^{n+1} - 2. \]

### Final Answer:
\[ \sum_{k=1}^{n} f(k) = 2^{n+1} - 2. \]