Suppose that $A$ is a Koszul ring (for the definition of Koszul ring see page 2 of BGS) and let $N$ be a graded $A$-module. As I understand it should be true that $$\operatorname{ext}_A ^{i+1} (A_0, N) = \hom_A (K^i,N)$$ where $K^i = \ker(P^i \rightarrow P^{i-1})$.
However I haven't been able to show it. The way I have been trying to show it is by induction since in the case were $i = 0$ maybe its possible to use the fact that $ext^1 _A (A_0, N) = \hom_A (K^0, N)/ Im(i^*)$ where $K^0 = \ker ( P^0 \rightarrow A_0)$ and $i: K^0 \hookrightarrow P^0$. But in order to even start here I have to show that $Im(i^*)$ is zero and I haven't been able to convince myself that it is.
I believe your question come from understanding proof of (3)$\Rightarrow$(1) of Prop 2.1.3, which actually require more than $N$ just to be graded module, it also require $N$ to be pure. Have you tried using that condition as well? –  Aaron Sep 3 '13 at 11:42