The remainder when $x^{100} -x^{99} +... -x +1$ is divided by $x^2 -1$ can be written in the form $ax +b$. Find $2a +b$.

[i]Proposed by Calvin Garces[/i]
1. Let the given polynomial be \( P(x) = x^{100} - x^{99} + x^{98} - \cdots - x + 1 \).
2. We need to find the remainder when \( P(x) \) is divided by \( x^2 - 1 \). Let the quotient be \( Q(x) \) and the remainder be \( ax + b \). Thus, we can write:
   \[
   P(x) = Q(x)(x^2 - 1) + ax + b
   \]
3. Since \( x^2 - 1 = (x - 1)(x + 1) \), we can evaluate \( P(x) \) at \( x = 1 \) and \( x = -1 \) to find the coefficients \( a \) and \( b \).
4. Evaluate \( P(x) \) at \( x = 1 \):
   \[
   P(1) = 1^{100} - 1^{99} + 1^{98} - \cdots - 1 + 1 = 1 - 1 + 1 - 1 + \cdots - 1 + 1 = 1
   \]
   Therefore:
   \[
   P(1) = a(1) + b = a + b = 1
   \]
5. Evaluate \( P(x) \) at \( x = -1 \):
   \[
   P(-1) = (-1)^{100} - (-1)^{99} + (-1)^{98} - \cdots - (-1) + 1 = 1 + 1 + 1 + \cdots + 1 = 101
   \]
   Therefore:
   \[
   P(-1) = a(-1) + b = -a + b = 101
   \]
6. We now have the system of linear equations:
   \[
   \begin{cases}
   a + b = 1 \\
   -a + b = 101
   \end{cases}
   \]
7. Solve the system of equations:
   \[
   \begin{aligned}
   a + b &= 1 \\
   -a + b &= 101
   \end{aligned}
   \]
   Add the two equations:
   \[
   (a + b) + (-a + b) = 1 + 101 \implies 2b = 102 \implies b = 51
   \]
   Substitute \( b = 51 \) into \( a + b = 1 \):
   \[
   a + 51 = 1 \implies a = 1 - 51 \implies a = -50
   \]
8. The remainder when \( P(x) \) is divided by \( x^2 - 1 \) is \( -50x + 51 \).
9. We need to find \( 2a + b \):
   \[
   2a + b = 2(-50) + 51 = -100 + 51 = -49
   \]

The final answer is \(\boxed{-49}\)