Easy spectral sequence 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.
 A: This is a typo in Weibel. The correct claim should be that $0 \to E_{0, n}^2 \to H_n \to E_{1, n - 1}^2 \to 0$ is exact.
You pretty much had it though. The convergence $E_{p, q}^n \Rightarrow H_{p + q}$ guarantees a filtration of $H_n$ for each $n$,
$$
0 = F_{-1} H_n \subseteq F_0 H_n \subseteq \cdots \subseteq F_{n - 1}H_n \subseteq F_n H_n = H_n
$$
such that $E_{p, q}^\infty \cong F_p H_{p + q} / F_{p - 1} H_{p + q}$.
The condition $E_{p, q}^2 = 0$ for $p \neq 0, 1$ tells us that 
$$
0 = F_{-1} H_n \subseteq F_0 H_n \subseteq F_1 H_n = \cdots = F_n H_n = H_n,
$$
since it would mean that $0 = E_{p, q}^2 = E_{p, q}^\infty$, for $p \neq 0, 1$, and in particular $E_{2, n - 2}^\infty \cong F_2 H_n / F_1 H_n = 0$, and inductively we get $F_1 H_n = \cdots = F_n H_n = H_n$.
Observe that $E_{0, n}^2 = E_{0, n}^\infty \cong F_0 H_n / F_{-1} H_n = F_0 H_n$ since $F_{-1} H_n = 0$. 
Now since $E_{1, n - 1}^2 = E_{1, n - 1}^\infty \cong F_1 H_n / F_0 H_n$ there is a short exact sequence 
$$
0 \to F_0 H_n \to F_1 H_n \to E_{1, n - 1}^\infty \to 0,
$$
which after plugging in the isomorphisms we've established, corresponds to,
$$
0 \to E_{0, n}^2 \to H_n \to E_{1, n - 1}^2 \to 0,
$$
as desired.
