Let `sum_(k=1)^(10)f(a+k)=16(2^(10)-1),` where the function f satisfies `f(x+y)=f(x)f(y)` for all natural numbers x, y and f(1) = 2. Then, the natural number 'a' is

Options

A. 2
B. 4
C. 3
D. 16

Explanation

To solve the problem, we need to find the natural number 'a' given the equation:

\[ \sum_{k=1}^{10} f(a+k) = 16(2^{10}-1) \]

where the function \( f \) satisfies \( f(x+y) = f(x)f(y) \) for all natural numbers \( x, y \) and \( f(1) = 2 \).

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

From the functional equation \( f(x+y) = f(x)f(y) \), we can deduce that \( f \) is an exponential function. Specifically, since \( f(1) = 2 \), we can express \( f(x) \) in the form:

\[ f(x) = f(1)^x = 2^x \]

### Step 2: Substitute \( f(x) \) into the summation

Now, we substitute \( f(x) \) into the summation:

\[ \sum_{k=1}^{10} f(a+k) = \sum_{k=1}^{10} 2^{a+k} = \sum_{k=1}^{10} 2^a \cdot 2^k = 2^a \sum_{k=1}^{10} 2^k \]

### Step 3: Calculate the summation \( \sum_{k=1}^{10} 2^k \)

The summation \( \sum_{k=1}^{10} 2^k \) is a geometric series where the first term \( a = 2 \) and the common ratio \( r = 2 \). The formula for the sum of the first \( n \) terms of a geometric series is:

\[ S_n = a \frac{r^n - 1}{r - 1} \]

Substituting \( a = 2 \), \( r = 2 \), and \( n = 10 \):

\[ \sum_{k=1}^{10} 2^k = 2 \frac{2^{10} - 1}{2 - 1} = 2(2^{10} - 1) = 2^{11} - 2 \]

### Step 4: Substitute back into the equation

Now substituting back into our equation:

\[ \sum_{k=1}^{10} f(a+k) = 2^a (2^{11} - 2) \]

We set this equal to the right side of the original equation:

\[ 2^a (2^{11} - 2) = 16(2^{10} - 1) \]

### Step 5: Simplify the right side

We can simplify the right side:

\[ 16(2^{10} - 1) = 16 \cdot 2^{10} - 16 = 2^4 \cdot 2^{10} - 16 = 2^{14} - 16 \]

### Step 6: Set the equations equal

Now we have:

\[ 2^a (2^{11} - 2) = 2^{14} - 16 \]

### Step 7: Factor out common terms

Factoring out \( 2 \) from the left side:

\[ 2^a (2^{11} - 2) = 2^{a+1}(2^{10} - 1) \]

Thus, we can rewrite the equation as:

\[ 2^{a+1}(2^{10} - 1) = 2^{14} - 16 \]

### Step 8: Set the powers of 2 equal

Now we can equate the powers of 2. We know that \( 2^{14} - 16 = 2^{14} - 2^4 = 2^4(2^{10} - 1) \). Thus:

\[ 2^{a+1}(2^{10} - 1) = 2^4(2^{10} - 1) \]

### Step 9: Solve for \( a \)

Since \( 2^{10} - 1 \) is not zero, we can divide both sides by \( 2^{10} - 1 \):

\[ 2^{a+1} = 2^4 \]

This implies:

\[ a + 1 = 4 \implies a = 3 \]

### Final Answer

The natural number \( a \) is:

\[ \boxed{3} \]