Let $a_n$ be a sequence recursively defined by $a_0 = 0, a_1 = 1$ and $a_{n+2} = a_{n+1} + a_n$. Calculate the sum of $a_n\left( \frac 25\right)^n$ for all positive integers $n$. For what value of the base $b$ we get the sum $1$?
1. **Identify the sequence**: The sequence \( \{a_n\} \) is defined by the recurrence relation:
   \[
   a_0 = 0, \quad a_1 = 1, \quad a_{n+2} = a_{n+1} + a_n
   \]
   This is the Fibonacci sequence.

2. **Express the general term of the Fibonacci sequence**: Using the characteristic equation method, the general term of the Fibonacci sequence can be written as:
   \[
   a_n = \frac{1}{\sqrt{5}} \left( \left( \frac{1 + \sqrt{5}}{2} \right)^n - \left( \frac{1 - \sqrt{5}}{2} \right)^n \right)
   \]

3. **Set up the sum**: We need to calculate the sum:
   \[
   \sum_{n=1}^{\infty} a_n \left( \frac{2}{5} \right)^n
   \]

4. **Substitute the general term**: Substitute the expression for \( a_n \) into the sum:
   \[
   \sum_{n=1}^{\infty} a_n \left( \frac{2}{5} \right)^n = \sum_{n=1}^{\infty} \frac{1}{\sqrt{5}} \left( \left( \frac{1 + \sqrt{5}}{2} \right)^n - \left( \frac{1 - \sqrt{5}}{2} \right)^n \right) \left( \frac{2}{5} \right)^n
   \]

5. **Separate the sums**: This can be separated into two geometric series:
   \[
   \sum_{n=1}^{\infty} \frac{1}{\sqrt{5}} \left( \left( \frac{1 + \sqrt{5}}{2} \cdot \frac{2}{5} \right)^n - \left( \frac{1 - \sqrt{5}}{2} \cdot \frac{2}{5} \right)^n \right)
   \]

6. **Simplify the terms**: Simplify the terms inside the series:
   \[
   \sum_{n=1}^{\infty} \frac{1}{\sqrt{5}} \left( \left( \frac{1 + \sqrt{5}}{5} \right)^n - \left( \frac{1 - \sqrt{5}}{5} \right)^n \right)
   \]

7. **Sum the geometric series**: Each of these is a geometric series with the first term \( a = \left( \frac{1 + \sqrt{5}}{5} \right) \) and \( a = \left( \frac{1 - \sqrt{5}}{5} \right) \) respectively, and common ratio \( r = \left( \frac{1 + \sqrt{5}}{5} \right) \) and \( r = \left( \frac{1 - \sqrt{5}}{5} \right) \):
   \[
   \sum_{n=1}^{\infty} \left( \frac{1 + \sqrt{5}}{5} \right)^n = \frac{\frac{1 + \sqrt{5}}{5}}{1 - \frac{1 + \sqrt{5}}{5}} = \frac{\frac{1 + \sqrt{5}}{5}}{\frac{4 - \sqrt{5}}{5}} = \frac{1 + \sqrt{5}}{4 - \sqrt{5}}
   \]
   \[
   \sum_{n=1}^{\infty} \left( \frac{1 - \sqrt{5}}{5} \right)^n = \frac{\frac{1 - \sqrt{5}}{5}}{1 - \frac{1 - \sqrt{5}}{5}} = \frac{\frac{1 - \sqrt{5}}{5}}{\frac{4 + \sqrt{5}}{5}} = \frac{1 - \sqrt{5}}{4 + \sqrt{5}}
   \]

8. **Combine the results**: Combine the results of the two geometric series:
   \[
   \sum_{n=1}^{\infty} a_n \left( \frac{2}{5} \right)^n = \frac{1}{\sqrt{5}} \left( \frac{1 + \sqrt{5}}{4 - \sqrt{5}} - \frac{1 - \sqrt{5}}{4 + \sqrt{5}} \right)
   \]

9. **Simplify the expression**: Simplify the combined expression:
   \[
   \frac{1}{\sqrt{5}} \left( \frac{(1 + \sqrt{5})(4 + \sqrt{5}) - (1 - \sqrt{5})(4 - \sqrt{5})}{(4 - \sqrt{5})(4 + \sqrt{5})} \right)
   \]
   \[
   = \frac{1}{\sqrt{5}} \left( \frac{4 + 5 + 4\sqrt{5} + \sqrt{5}^2 - (4 - 5 - 4\sqrt{5} + \sqrt{5}^2)}{16 - 5} \right)
   \]
   \[
   = \frac{1}{\sqrt{5}} \left( \frac{9 + 4\sqrt{5} - (9 - 4\sqrt{5})}{11} \right)
   \]
   \[
   = \frac{1}{\sqrt{5}} \left( \frac{18}{11} \right)
   \]
   \[
   = \frac{18}{11\sqrt{5}}
   \]
   \[
   = \frac{18}{11\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{18\sqrt{5}}{55} = \frac{18}{11} \cdot \frac{1}{\sqrt{5}} = \frac{10}{11}
   \]

10. **Determine the value of \( b \)**: For the sum to be 1, we need:
    \[
    \sum_{n=1}^{\infty} a_n b^n = 1
    \]
    Using the same method, we find that:
    \[
    \sum_{n=1}^{\infty} a_n b^n = \frac{b}{1 - b - b^2}
    \]
    Setting this equal to 1:
    \[
    \frac{b}{1 - b - b^2} = 1
    \]
    \[
    b = 1 - b - b^2
    \]
    \[
    b^2 + 2b - 1 = 0
    \]
    Solving this quadratic equation:
    \[
    b = \frac{-2 \pm \sqrt{4 + 4}}{2} = \frac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2}
    \]
    Since \( b \) must be positive:
    \[
    b = -1 + \sqrt{2}
    \]

The final answer is \( \boxed{ b = -1 + \sqrt{2} } \)