`f:N to R` be a function such that `f(x+y)=f(x).f(y)` for natural number `x & y` if `f(1)=2 & sum_(k=1)^20 f(a+k)=512(2^20-1)` then the value of `a` is

Answer

To solve the problem, we need to find the value of \( a \) given the function \( f: \mathbb{N} \to \mathbb{R} \) defined by the functional equation \( f(x+y) = f(x) \cdot f(y) \) for natural numbers \( x \) and \( y \), with the condition \( f(1) = 2 \) and the summation condition \( \sum_{k=1}^{20} f(a+k) = 512(2^{20}-1) \).

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

From the functional equation \( f(x+y) = f(x) \cdot f(y) \), we can deduce that \( f(n) \) has an exponential form. Specifically, we can express \( f(n) \) in terms of \( f(1) \):

1. For \( n = 1 \):
\[ f(1) = 2 \]

2. For \( n = 2 \):
\[ f(2) = f(1+1) = f(1) \cdot f(1) = 2 \cdot 2 = 4 \]

3. For \( n = 3 \):
\[ f(3) = f(2+1) = f(2) \cdot f(1) = 4 \cdot 2 = 8 \]

4. Continuing this pattern, we can generalize:
\[ f(n) = f(1)^n = 2^n \]

Thus, we conclude that:
\[ f(n) = 2^n \]

### Step 2: Substitute into the summation condition

Now, we substitute \( f(n) = 2^n \) into the summation condition:
\[ \sum_{k=1}^{20} f(a+k) = \sum_{k=1}^{20} 2^{a+k} \]

This can be rewritten as:
\[ \sum_{k=1}^{20} 2^{a+k} = 2^a \sum_{k=1}^{20} 2^k \]

### Step 3: Calculate the sum \( \sum_{k=1}^{20} 2^k \)

The sum \( \sum_{k=1}^{20} 2^k \) is a geometric series:
\[ \sum_{k=1}^{20} 2^k = 2(1 + 2 + 2^2 + \ldots + 2^{19}) = 2 \cdot \frac{2^{20} - 1}{2 - 1} = 2(2^{20} - 1) = 2^{21} - 2 \]

### Step 4: Substitute back into the equation

Now substituting this back into our equation:
\[ 2^a (2^{21} - 2) = 512(2^{20} - 1) \]

### Step 5: Simplify the equation

We know that \( 512 = 2^9 \), so we can rewrite the equation:
\[ 2^a (2^{21} - 2) = 2^9 (2^{20} - 1) \]

Expanding both sides:
\[ 2^a (2^{21} - 2) = 2^a \cdot 2 \cdot (2^{20} - 1) \]
\[ 2^{a+1} (2^{20} - 1) = 2^9 (2^{20} - 1) \]

### Step 6: Cancel out common terms

Assuming \( 2^{20} - 1 \neq 0 \), we can divide both sides by \( 2^{20} - 1 \):
\[ 2^{a+1} = 2^9 \]

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

This implies:
\[ a + 1 = 9 \implies a = 8 \]

### Final Answer:
Thus, the value of \( a \) is:
\[ \boxed{8} \]