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 .
If ` sum_(k=1)^(n)f(a+k)=16(2^(n)-1) `, then a=

Options

A. 3
B. 4
C. 2
D. none of these

Explanation

To solve the problem, we need to find the value of \( a \) given the functional equation and the summation condition. Let's go through the steps systematically.

### Step 1: Determine the function \( f(x) \)

We are given the functional equation:

\[ f(x+y) = f(x)f(y) \]

for all \( x, y \in \mathbb{R} \) and that \( f(1) = 2 \).

### Step 2: Find values of \( f \) at specific points

1. **Calculate \( f(2) \)**:
Set \( x = 1 \) and \( y = 1 \):
\[ f(2) = f(1+1) = f(1)f(1) = 2 \cdot 2 = 4 \]

2. **Calculate \( f(3) \)**:
Set \( x = 2 \) and \( y = 1 \):
\[ f(3) = f(2+1) = f(2)f(1) = 4 \cdot 2 = 8 \]

3. **Calculate \( f(4) \)**:
Set \( x = 2 \) and \( y = 2 \):
\[ f(4) = f(2+2) = f(2)f(2) = 4 \cdot 4 = 16 \]

From the calculations, we observe a pattern:
- \( f(1) = 2 = 2^1 \)
- \( f(2) = 4 = 2^2 \)
- \( f(3) = 8 = 2^3 \)
- \( f(4) = 16 = 2^4 \)

### Step 3: Generalize the function

From the pattern, we can conjecture that:
\[ f(n) = 2^n \quad \text{for all } n \in \mathbb{N} \]

### Step 4: Verify the function

To verify, we can check if \( f(x) = 2^x \) satisfies the original functional equation:
\[ f(x+y) = 2^{x+y} = 2^x \cdot 2^y = f(x)f(y) \]
Thus, \( f(x) = 2^x \) is indeed a solution.

### Step 5: Use the summation condition

We need to evaluate:
\[ \sum_{k=1}^{n} f(a+k) = 16(2^n - 1) \]

Substituting \( f(x) = 2^x \):
\[ \sum_{k=1}^{n} f(a+k) = \sum_{k=1}^{n} 2^{a+k} = \sum_{k=1}^{n} 2^a \cdot 2^k = 2^a \sum_{k=1}^{n} 2^k \]

The sum \( \sum_{k=1}^{n} 2^k \) is a geometric series:
\[ \sum_{k=1}^{n} 2^k = 2(2^n - 1) \]

Thus:
\[ \sum_{k=1}^{n} f(a+k) = 2^a \cdot 2(2^n - 1) = 2^{a+1}(2^n - 1) \]

### Step 6: Set the equation equal to the given condition

We set this equal to the given condition:
\[ 2^{a+1}(2^n - 1) = 16(2^n - 1) \]

### Step 7: Solve for \( a \)

Dividing both sides by \( (2^n - 1) \) (assuming \( n \geq 1 \)):
\[ 2^{a+1} = 16 \]

Since \( 16 = 2^4 \):
\[ a + 1 = 4 \implies a = 3 \]

### Conclusion

The value of \( a \) is:
\[ \boxed{3} \]