This is a question in Weibel's Homological Algebra.
Suppose that a spectral sequence converging to $H_\ast$ has $E^2_{pq} = 0$ unless $p = 0,1$. Show that there are exact sequences $0 \rightarrow E^2_{1,n-1} \rightarrow H_n \rightarrow E^2_{0,n} \rightarrow 0.$
And what I have done: It is easy to see that $E^\infty_{pq} = E^2_{pq}$. It is also not hard to prove that each $H_n$ must have a filtration of the form $0 = F_{-1}H_n \subseteq F_0H_n \subseteq F_1H_n = H_n.$
It then follow that $F_0H_n/F_{-1}H_n = F_0H_n = E^2_{0n}$ and $F_1H_n/F_0H_n = H_n/E^2_{0n} = E^2_{1,n-1}$. But these give maps $H_n \rightarrow E^2_{1,n-1}$ and $E^2_{0n} \subseteq H_n$. I don't know how to get the maps in the other direction.