Consider the category of chain complexes over a ring $R$.


We can show that $\text{Ext}^1(M, N)$ classifies extensions using the triangulated category structure: the homotopy kernel of a map $N \rightarrow M[1]$, $K$, gives rise to the designated triangle $M \rightarrow K \rightarrow N$. This gives a simple picture of Yoneda's extension theorem for $\text{Ext}^1(M, N)$, which also seems to rely on the "threeness" of a triangulated category.


My question is, is there an extension of this view to $\text{Ext}^n(M, N)$?


The proof of the situation for $\text{Ext}^1$ is quite simple using triangulated categories and it would be nice if the situation could be handled the same way for $\text{Ext}^n$.



Yoneda Ext theorem and extensions
We have that $\mathrm{Ext}^n(M, N) = \mathrm{Hom}(M, N[n])$, in other words, an element of $\mathrm{Ext}^n(M, N)$ is a map
$$
M \longrightarrow I\_N[n]
$$
where $N \to I\_N$ is an injective resolution. From this we obtain a distinguished triangle,
$$
M \longrightarrow I\_N[n] \longrightarrow C \overset{+}\longrightarrow
$$
The homotopical invariant gives an extension of complexes
$$
0 \longrightarrow I\_N[n] \longrightarrow C \longrightarrow M \longrightarrow 0 \
$$
which, plugin-in the beginning of the resolution of $N$, yields the classical $n$-extension
$$
0 \longrightarrow N \longrightarrow I^0\_N \longrightarrow I^1\_N \cdots \longrightarrow I^{n-1}\_N \longrightarrow M \longrightarrow 0
$$
as wanted


