The sequence an is defined by $a_0 = 4, a_1 = 1$ and the recurrence formula $a_{n+1} = a_n + 6a_{n-1}$. The sequence $b_n$ is given by
\[b_n=\sum_{k=0}^n \binom nk a_k.\]
Find the coefficients $\alpha,\beta$  so that $b_n$ satisfies the recurrence formula $b_{n+1} = \alpha b_n + \beta b_{n-1}$. Find the explicit form of $b_n$.
1. **Define the sequence \(a_n\)**:
   The sequence \(a_n\) is given by:
   \[
   a_0 = 4, \quad a_1 = 1, \quad \text{and} \quad a_{n+1} = a_n + 6a_{n-1}
   \]

2. **Find the characteristic equation**:
   The recurrence relation \(a_{n+1} = a_n + 6a_{n-1}\) has the characteristic equation:
   \[
   x^2 - x - 6 = 0
   \]
   Solving this quadratic equation, we get the roots:
   \[
   x = \frac{1 \pm \sqrt{1 + 24}}{2} = \frac{1 \pm 5}{2}
   \]
   Thus, the roots are:
   \[
   r_1 = 3, \quad r_2 = -2
   \]

3. **General solution for \(a_n\)**:
   The general solution for the recurrence relation is:
   \[
   a_n = A \cdot 3^n + B \cdot (-2)^n
   \]
   Using the initial conditions \(a_0 = 4\) and \(a_1 = 1\), we can solve for \(A\) and \(B\):
   \[
   a_0 = A \cdot 3^0 + B \cdot (-2)^0 = A + B = 4
   \]
   \[
   a_1 = A \cdot 3^1 + B \cdot (-2)^1 = 3A - 2B = 1
   \]
   Solving these equations simultaneously:
   \[
   A + B = 4
   \]
   \[
   3A - 2B = 1
   \]
   Multiply the first equation by 2:
   \[
   2A + 2B = 8
   \]
   Add this to the second equation:
   \[
   3A - 2B + 2A + 2B = 1 + 8
   \]
   \[
   5A = 9 \implies A = \frac{9}{5}
   \]
   Substitute \(A = \frac{9}{5}\) into \(A + B = 4\):
   \[
   \frac{9}{5} + B = 4 \implies B = 4 - \frac{9}{5} = \frac{20}{5} - \frac{9}{5} = \frac{11}{5}
   \]
   Thus, the explicit form of \(a_n\) is:
   \[
   a_n = \frac{9}{5} \cdot 3^n + \frac{11}{5} \cdot (-2)^n
   \]

4. **Define the sequence \(b_n\)**:
   The sequence \(b_n\) is given by:
   \[
   b_n = \sum_{k=0}^n \binom{n}{k} a_k
   \]

5. **Find the explicit form of \(b_n\)**:
   Using the binomial theorem, we can write:
   \[
   b_n = \sum_{k=0}^n \binom{n}{k} \left( \frac{9}{5} \cdot 3^k + \frac{11}{5} \cdot (-2)^k \right)
   \]
   This can be split into two sums:
   \[
   b_n = \frac{9}{5} \sum_{k=0}^n \binom{n}{k} 3^k + \frac{11}{5} \sum_{k=0}^n \binom{n}{k} (-2)^k
   \]
   Using the binomial theorem:
   \[
   \sum_{k=0}^n \binom{n}{k} 3^k = (1 + 3)^n = 4^n
   \]
   \[
   \sum_{k=0}^n \binom{n}{k} (-2)^k = (1 - 2)^n = (-1)^n
   \]
   Therefore:
   \[
   b_n = \frac{9}{5} \cdot 4^n + \frac{11}{5} \cdot (-1)^n
   \]

6. **Find the recurrence relation for \(b_n\)**:
   We need to find \(\alpha\) and \(\beta\) such that:
   \[
   b_{n+1} = \alpha b_n + \beta b_{n-1}
   \]
   Using the explicit form of \(b_n\):
   \[
   b_{n+1} = \frac{9}{5} \cdot 4^{n+1} + \frac{11}{5} \cdot (-1)^{n+1}
   \]
   \[
   b_{n+1} = \frac{9}{5} \cdot 4 \cdot 4^n + \frac{11}{5} \cdot (-1) \cdot (-1)^n
   \]
   \[
   b_{n+1} = \frac{36}{5} \cdot 4^n + \frac{11}{5} \cdot (-1)^{n+1}
   \]
   Comparing this with \(\alpha b_n + \beta b_{n-1}\):
   \[
   \alpha \left( \frac{9}{5} \cdot 4^n + \frac{11}{5} \cdot (-1)^n \right) + \beta \left( \frac{9}{5} \cdot 4^{n-1} + \frac{11}{5} \cdot (-1)^{n-1} \right)
   \]
   We need to match coefficients:
   \[
   \alpha \cdot \frac{9}{5} \cdot 4^n + \beta \cdot \frac{9}{5} \cdot 4^{n-1} = \frac{36}{5} \cdot 4^n
   \]
   \[
   \alpha \cdot \frac{11}{5} \cdot (-1)^n + \beta \cdot \frac{11}{5} \cdot (-1)^{n-1} = \frac{11}{5} \cdot (-1)^{n+1}
   \]
   Solving these, we get:
   \[
   \alpha \cdot \frac{9}{5} = \frac{36}{5} \implies \alpha = 4
   \]
   \[
   \alpha \cdot \frac{11}{5} + \beta \cdot \frac{11}{5} \cdot (-1) = \frac{11}{5} \cdot (-1) \implies \alpha - \beta = -1
   \]
   \[
   4 - \beta = -1 \implies \beta = 5
   \]
   Thus, the recurrence relation for \(b_n\) is:
   \[
   b_{n+1} = 4b_n + 5b_{n-1}
   \]

The final answer is \( \boxed{ b_n = \frac{9}{5} \cdot 4^n + \frac{11}{5} \cdot (-1)^n } \)