Let $\{u_n \}$ be the sequence defined by its first two terms $u_0, u_1$  and the recursion formula
\[u_{n+2 }= u_n - u_{n+1}.\]
[b](a)[/b] Show that $u_n$ can be written in the form $u_n = \alpha a^n + \beta b^n$, where $a, b, \alpha, \beta$ are constants independent of $n$ that have to be determined.

[b](b)[/b] If $S_n = u_0 + u_1 + \cdots + u_n$, prove that $S_n + u_{n-1}$ is a constant independent of $n.$ Determine this constant.
**(a)** Show that \( u_n \) can be written in the form \( u_n = \alpha a^n + \beta b^n \), where \( a, b, \alpha, \beta \) are constants independent of \( n \) that have to be determined.

1. Consider the recurrence relation \( u_{n+2} = u_n - u_{n+1} \).
2. To solve this, we look for solutions of the form \( u_n = r^n \).
3. Substituting \( u_n = r^n \) into the recurrence relation, we get:
   \[
   r^{n+2} = r^n - r^{n+1}
   \]
4. Dividing through by \( r^n \) (assuming \( r \neq 0 \)), we obtain the characteristic equation:
   \[
   r^2 = 1 - r
   \]
5. Rearranging, we get the quadratic equation:
   \[
   r^2 + r - 1 = 0
   \]
6. Solving this quadratic equation using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 1 \), and \( c = -1 \), we get:
   \[
   r = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2}
   \]
7. Thus, the roots are:
   \[
   a = \frac{-1 + \sqrt{5}}{2}, \quad b = \frac{-1 - \sqrt{5}}{2}
   \]
8. Therefore, the general solution to the recurrence relation is:
   \[
   u_n = \alpha a^n + \beta b^n
   \]
   where \( \alpha \) and \( \beta \) are constants determined by the initial conditions \( u_0 \) and \( u_1 \).

**(b)** If \( S_n = u_0 + u_1 + \cdots + u_n \), prove that \( S_n + u_{n-1} \) is a constant independent of \( n \). Determine this constant.

1. Define \( S_n = u_0 + u_1 + \cdots + u_n \).
2. Consider \( S_{n+1} \):
   \[
   S_{n+1} = S_n + u_{n+1}
   \]
3. We need to show that \( S_n + u_{n-1} \) is constant. Consider:
   \[
   S_{n+1} + u_n = (S_n + u_{n+1}) + u_n
   \]
4. Using the recurrence relation \( u_{n+2} = u_n - u_{n+1} \), we can rewrite \( u_{n+1} \) as:
   \[
   u_{n+1} = u_{n+2} + u_{n+1}
   \]
5. Thus:
   \[
   S_{n+1} + u_n = S_n + u_{n+1} + u_n = S_n + u_{n-1} + (u_{n+1} + u_n - u_{n-1})
   \]
6. Simplifying, we get:
   \[
   S_{n+1} + u_n = S_n + u_{n-1}
   \]
7. Therefore, \( S_n + u_{n-1} \) is indeed a constant independent of \( n \).
8. To determine this constant, consider \( n = 1 \):
   \[
   S_1 = u_0 + u_1
   \]
   \[
   S_1 + u_0 = u_0 + u_1 + u_0 = 2u_0 + u_1
   \]
9. Hence, the constant is:
   \[
   2u_0 + u_1
   \]

The final answer is \( \boxed{ 2u_0 + u_1 } \)