Exact sequence associated with a normal series

For a normal series of finite length
Suppose $$G$$ is a group with a normal series:

$$\{ e \} = K_0 \le K_1 \le K_2 \le \dots \le K_n = G$$

This means that each $$K_i$$ is a normal subgroup of $$G$$.

There is an exact sequence of groups defined as follows:

$$\{ e \} = K_0 \to K_1 \to K_2 \to K_3/K_1 \to K_4/K_2 \to \dots K_i/K_{i-2} \to \dots \to K_n/K_{n-2} \to K_n/K_{n-1} \to \{ e \}$$

where the first two maps are inclusions, the lat two maps are the natural quotient maps, and the other maps are obtained by composing an inclusion with a quotient map. The maps are exact: the image of the map going to $$K_i/K_{i-2}$$ equals the kernel of the map going from $$K_i/K_{i-2}$$, and both are equal to $$K_{i-1}/K_{i-2}$$.

The ends look unnatural, but we can make them natural by setting $$K_{-2} = K_{-1} = K_0$$ and setting $$K_{n+1} = K_{n+2} = K_n$$.

Note that we can weaken somewhat the assumption that $$K_i$$ are all normal in $$G$$. What we really need is a 2-step subnormal series: a subnormal series where $$K_{i-2}$$ is normal in $$K_i$$ for each $$i$$.