# The module $M/M_n$ is of finite length

I have this doubt from a proposition in Dimension theory from Atiyah Macdonald. The proposition is as follows.

Let $A$ be a noetherian local ring, $\mathcal{m}$ be its maximal ideal, $\mathcal{q}$ an $\mathcal{m}$-primary ideal, $M$ a finitely generated $A$-module, $(M_n)$ a stable $\mathcal{q}$-filtration of $M$. Then, $M/M_n$ has finite length for each $n\geq 0$.

The proof goes as follows. Let $G(A)=\oplus_n\mathcal{q}/\mathcal{q}^{n+1}$, $G(M)=\oplus_m M_n/M_{n+1}$. Then $G_0(A)=A/\mathcal{q}$ is an artin local ring, $G(A)$ is noetherian and $G(M)$ is a finitely generated graded $G(A)$ module. Each $G_n(M)=M_n/M_{n+1}$ is a Notherian $A$-module annihilated by $\mathcal{q}$, hence a noetherian $A/\mathcal{q}$ module and therefore is of finite length since $A/\mathcal{q}$ is Artin. So far it is okay, from this they conclude saying

Hence $M/M_n$ is of finite length and $l(M/M_n)=\Sigma_{r=1}^n l(M_r/M_{r+1})$. How do we get this. I am not able to find any exact sequence where I can use the additivity of length. For example, I thought for $n=2$, this would work, $0\longrightarrow M/M_1\longrightarrow M/M_2\longrightarrow M_1/M_2\longrightarrow 0$. But I am not even able to define maps to make this an SES. ANy help will be appreciated.

Use the short exact sequence $0\rightarrow M_1/M_2 \rightarrow M/M_2 \rightarrow M/M_1 \rightarrow 0$, where all homomorphisms are naturally defined.