Generalised Mayer-Vietoris long exact sequence In chapter 8 of Bott/Tu, the authors generalise the standard Mayer-Vietoris sequence to the setting of a countable open cover of $X$. Let's fix a countable cover $\{ U_i\}$ of $X$. According to Prop 8.5 of Bott/Tu, the sequence
$$ 0 \rightarrow \Omega^*(X) \rightarrow \bigoplus_{i} \Omega^*(U_i) \rightarrow \bigoplus_{i<j} \Omega^*(U_{ij}) \rightarrow \bigoplus_{i<j<k} \Omega^*(U_{ijk}) \rightarrow \cdots  $$
is exact. Later in this chapter they use this sequence to prove some essential properties of the de Rham-Cech bicomplex, and then move into a study of spectral sequences of bicomplexes.
I'm trying to understand this sequence for the simplest case of three covering sets, say $U_1, U_2$ and $U_3$. In this case, the above sequence terminates quickly:
$$ 0 \rightarrow \Omega^*(X) \rightarrow \bigoplus_{i} \Omega^*(U_i) \rightarrow \bigoplus_{i<j} \Omega^*(U_{ij}) \rightarrow \Omega^*(U_{123}) \rightarrow 0  $$
My question is this: is there a long exact sequence for 3 sets that generalises the usual binary Mayer-Vietoris sequence? A naiive guess would be something like:
$$ \cdots \rightarrow H^q(U_{123})\rightarrow H^{q+1}(X)\rightarrow \bigoplus_i H^{q+1}(U_i) \rightarrow \bigoplus_{i<j} H^{q+1}(U_{ij}) \rightarrow H^{q+1}(U_{123}) \rightarrow \cdots $$
Almost all of these maps can be obtained by descending the associated maps that come from the exact sequence of differential forms. I would guess that the connecting homomorphism can be somehow obtained from the associated spectral sequence, but I am not too sure about this.

If you cover the circle by three intervals that pairwise intersect but not mutually, then $U_{123}$ is empty, $X = S^1$ and $U_i$ are contractible. So you cannot have an exact sequence of the form you guess, as at $H^1(X)$ you'd get
$$0 \to \mathbb{R} \to 0$$
which cannot be exact.
As @Mariano describes in the comments, the Mayer-Vietoris spectral sequence is the right generalization: it reduces to the usual exact sequence if the cover has only $2$ terms.
Alternatively you could also use an iterated exact sequence by first writing down exact sequences for $U_1 \cup U_2$ and $(U_1 \cup U_2) \cap U_3 = U_{13} \cup U_{23}$ and then for $X = (U_1 \cup U_2) \cup U_3$. This is equivalent to the spectral sequence approach, but breaks symmetry between the indices.
