Let $(a_n)$ be defined by $a_1=a_2=1$ and $a_n=a_{n-1}+a_{n-2}$ for $n>2$. Compute the sum $\frac{a_1}2+\frac{a_2}{2^2}+\frac{a_3}{2^3}+\ldots$.

1. Let \( S = \frac{a_1}{2} + \frac{a_2}{2^2} + \frac{a_3}{2^3} + \ldots \). Given the sequence \( (a_n) \) defined by \( a_1 = a_2 = 1 \) and \( a_n = a_{n-1} + a_{n-2} \) for \( n > 2 \), we recognize that \( (a_n) \) is the Fibonacci sequence.

2. We can express \( S \) as:
   \[
   S = \frac{1}{2} + \frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \ldots
   \]

3. To simplify, consider \( 2S \):
   \[
   2S = 1 + \frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \ldots
   \]

4. Subtract \( S \) from \( 2S \):
   \[
   2S - S = S = 1 + \left( \frac{1}{2} - \frac{1}{2} \right) + \left( \frac{2}{4} - \frac{1}{4} \right) + \left( \frac{3}{8} - \frac{2}{8} \right) + \ldots
   \]
   Simplifying the terms inside the parentheses:
   \[
   S = 1 + 0 + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \ldots
   \]

5. Notice that the remaining series is a geometric series with the first term \( \frac{1}{4} \) and common ratio \( \frac{1}{2} \):
   \[
   \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \ldots = \sum_{k=2}^{\infty} \frac{1}{2^k}
   \]

6. The sum of an infinite geometric series \( \sum_{k=0}^{\infty} ar^k \) is given by \( \frac{a}{1-r} \), where \( a \) is the first term and \( r \) is the common ratio. Here, \( a = \frac{1}{4} \) and \( r = \frac{1}{2} \):
   \[
   \sum_{k=2}^{\infty} \frac{1}{2^k} = \frac{\frac{1}{4}}{1 - \frac{1}{2}} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{4} \cdot 2 = \frac{1}{2}
   \]

7. Therefore, we have:
   \[
   S = 1 + \frac{1}{2}
   \]

8. Solving for \( S \):
   \[
   S = 1 + \frac{1}{2} = \frac{2}{2} + \frac{1}{2} = \frac{3}{2}
   \]

9. However, this contradicts our earlier steps. Let's re-evaluate the subtraction step:
   \[
   S = 1 + \frac{S}{2}
   \]
   Solving for \( S \):
   \[
   S - \frac{S}{2} = 1 \implies \frac{S}{2} = 1 \implies S = 2
   \]

The final answer is \(\boxed{2}\)