Let `f:N rarr R` be a function such that f(x+y)=2f(x) f(y) for natural numbers x and y. If f(1)=2, then the value of a for which `sum_(k=1)^(10)f(alpha+k)=512/3(2^(20)-1)`

Options

A. 2
B. 3
C. 4
D. 6

Answer

To solve the problem step by step, we need to analyze the given functional equation and the conditions provided.

### Step 1: Understand the functional equation
We have the function \( f: \mathbb{N} \to \mathbb{R} \) defined by the equation:
\[ f(x+y) = 2f(x)f(y) \]
for natural numbers \( x \) and \( y \). We also know that \( f(1) = 2 \).

### Step 2: Find \( f(2) \)
Using the functional equation, we can find \( f(2) \):
\[ f(2) = f(1+1) = 2f(1)f(1) = 2 \cdot 2 \cdot 2 = 8 \]

### Step 3: Find \( f(3) \)
Next, we calculate \( f(3) \):
\[ f(3) = f(2+1) = 2f(2)f(1) = 2 \cdot 8 \cdot 2 = 32 \]

### Step 4: Find \( f(4) \)
Now, we find \( f(4) \):
\[ f(4) = f(3+1) = 2f(3)f(1) = 2 \cdot 32 \cdot 2 = 128 \]

### Step 5: Find \( f(5) \)
Continuing, we find \( f(5) \):
\[ f(5) = f(4+1) = 2f(4)f(1) = 2 \cdot 128 \cdot 2 = 512 \]

### Step 6: Find \( f(6) \)
Next, we calculate \( f(6) \):
\[ f(6) = f(5+1) = 2f(5)f(1) = 2 \cdot 512 \cdot 2 = 2048 \]

### Step 7: Find \( f(7) \)
Now, we find \( f(7) \):
\[ f(7) = f(6+1) = 2f(6)f(1) = 2 \cdot 2048 \cdot 2 = 8192 \]

### Step 8: Find \( f(8) \)
Next, we calculate \( f(8) \):
\[ f(8) = f(7+1) = 2f(7)f(1) = 2 \cdot 8192 \cdot 2 = 32768 \]

### Step 9: Find \( f(9) \)
Now, we find \( f(9) \):
\[ f(9) = f(8+1) = 2f(8)f(1) = 2 \cdot 32768 \cdot 2 = 131072 \]

### Step 10: Find \( f(10) \)
Finally, we calculate \( f(10) \):
\[ f(10) = f(9+1) = 2f(9)f(1) = 2 \cdot 131072 \cdot 2 = 524288 \]

### Step 11: Generalize \( f(n) \)
From the calculations, we observe that:
- \( f(1) = 2 = 2^1 \)
- \( f(2) = 8 = 2^3 \)
- \( f(3) = 32 = 2^5 \)
- \( f(4) = 128 = 2^7 \)
- \( f(5) = 512 = 2^9 \)
- \( f(6) = 2048 = 2^{11} \)
- \( f(7) = 8192 = 2^{13} \)
- \( f(8) = 32768 = 2^{15} \)
- \( f(9) = 131072 = 2^{17} \)
- \( f(10) = 524288 = 2^{19} \)

We can see that:
\[ f(n) = 2^{2n - 1} \]

### Step 12: Calculate \( \sum_{k=1}^{10} f(\alpha + k) \)
Now, we need to calculate:
\[ \sum_{k=1}^{10} f(\alpha + k) = \sum_{k=1}^{10} 2^{2(\alpha + k) - 1} = \sum_{k=1}^{10} 2^{2\alpha + 2k - 1} \]
This can be simplified as:
\[ = 2^{2\alpha - 1} \sum_{k=1}^{10} 2^{2k} = 2^{2\alpha - 1} \cdot (2^2 + 2^4 + 2^6 + 2^8 + 2^{10} + 2^{12} + 2^{14} + 2^{16} + 2^{18} + 2^{20}) \]

### Step 13: Sum the geometric series
The sum \( S = 2^2 + 2^4 + 2^6 + \ldots + 2^{20} \) is a geometric series with:
- First term \( a = 4 \) (which is \( 2^2 \))
- Common ratio \( r = 4 \)
- Number of terms \( n = 10 \)

The sum of a geometric series is given by:
\[ S_n = a \frac{r^n - 1}{r - 1} = 4 \frac{4^{10} - 1}{4 - 1} = \frac{4(2^{20} - 1)}{3} \]

### Step 14: Combine results
Thus, we have:
\[ \sum_{k=1}^{10} f(\alpha + k) = 2^{2\alpha - 1} \cdot \frac{4(2^{20} - 1)}{3} \]

### Step 15: Set the equation
We need to set this equal to \( \frac{512}{3}(2^{20} - 1) \):
\[ 2^{2\alpha - 1} \cdot \frac{4(2^{20} - 1)}{3} = \frac{512}{3}(2^{20} - 1) \]

### Step 16: Solve for \( \alpha \)
Cancel \( \frac{1}{3}(2^{20} - 1) \) from both sides:
\[ 2^{2\alpha - 1} \cdot 4 = 512 \]
This simplifies to:
\[ 2^{2\alpha + 1} = 512 \]
Since \( 512 = 2^9 \), we have:
\[ 2\alpha + 1 = 9 \implies 2\alpha = 8 \implies \alpha = 4 \]

### Final Answer
Thus, the value of \( \alpha \) is:
\[ \alpha = 4 \]