= Solution
Fix a finite-dimensional nonnegative <filtration of a ring> giving $R$ its <almost commutative algebra> structure, and choose a <good filtration of a module> on $M$. When $\operatorname{gr}R$ is a <standard graded algebra>, the <Hilbert-Serre theorem> makes the cumulative Hilbert function $h_M(j)=\dim_k F_jM$ eventually a <polynomial>. For $M\ne0$, define the <dimension of a filtered module> and <multiplicity of a filtered module> by
$$
\boxed{d(M)=\deg h_M,\qquad
m(M)=d(M)!\,[j^{d(M)}]h_M(j)}.
$$
Use $d(0)=-\infty$ and $m(0)=0$. With positive weighted generators instead of a standard grading, the cumulative Hilbert function is eventually a <quasipolynomial> with a common positive leading coefficient across residue classes. Equivalently, $h_M(j)=c j^d+O(j^{d-1})$, and the definitions are $d(M)=d$ and $m(M)=d!\,c$. The common leading coefficient follows from monotonicity: interlacing consecutive residue classes forces their leading coefficients to agree. For a standard grading, multiplicity is a positive integer, since an integer-valued <polynomial> has an integral expansion in the binomial polynomials; weighted normalization can instead give a positive rational number.
Two <good filtrations of a module> for this fixed algebra filtration bound each other after fixed index shifts. The corresponding Hilbert functions therefore satisfy $h_F(j-a)\leq h_G(j)\leq h_F(j+b)$ for fixed $a,b$. Comparing their leading growth shows that the degree and leading coefficient agree. Thus $d(M)$ and $m(M)$ are independent of the module filtration; the fixed algebra filtration is part of the multiplicity convention.
For a <submodule> $N$, use its <subspace filtration> and the <quotient filtration> on $M/N$. These are good because $\operatorname{gr}R$ is <Noetherian>. The <short exact sequence> of filtered pieces gives
$$
h_M(j)=h_N(j)+h_{M/N}(j).
$$
Nonzero leading coefficients are positive, so
$$
\boxed{d(M)=\max\{d(N),d(M/N)\}}.
$$
If $d(N)=d(M/N)$, those leading coefficients add and
$$
\boxed{m(M)=m(N)+m(M/N)}.
$$
This is <dimension and multiplicity in a filtered exact sequence>.
For the <Weyl algebra> $A_n(k)$ in characteristic zero, <Bernstein inequality for Weyl algebra modules> asserts $d(M)\geq n$ for every nonzero <finitely generated module> $M$. Write its generators as $x_i,D_i$, with $[D_i,x_j]=\delta_{ij}$ and all other generator commutators zero. The ordered monomials $x^\alpha D^\beta$ give the <Bernstein filtration>
$$
A_j=\operatorname{span}_k\{x^\alpha D^\beta:|\alpha|+|\beta|\leq j\},
\qquad\dim_k A_j=\binom{j+2n}{2n}.
$$
Take a finite-dimensional generating subspace $V\ne0$ of $M$ and put $M_j=A_jV$.
To prove <faithful finite-step action of a Weyl algebra>, consider $0\ne a\in A_j$ and choose a nonzero coefficient $c_{\alpha\beta}$ of a monomial of maximal total degree $d\leq j$. Applying $\prod_i(\operatorname{ad}D_i)^{\alpha_i}(\operatorname{ad}x_i)^{\beta_i}$ to $a$ gives
$$
\lambda=c_{\alpha\beta}\alpha!\,\beta!\,(-1)^{|\beta|}\ne0.
$$
Indeed, these <commutators> differentiate the ordered coefficients: lower-degree monomials vanish, and another degree-$d$ monomial can survive only if all its exponents dominate $(\alpha,\beta)$, which forces equality. This is <scalar extraction by iterated Weyl commutators>.
Expanding these iterated <commutators> writes $\lambda=\sum_\ell u_\ell a v_\ell$ with $u_\ell,v_\ell\in A_j$. If $a$ annihilated $M_j$, every summand would annihilate $V$, contrary to $\lambda V\ne0$. Therefore the action map is <injective>:
$$
A_j\hookrightarrow\operatorname{Hom}_k(M_j,M_{2j}).
$$
Taking dimensions and comparing polynomial degrees gives
$$
\binom{j+2n}{2n}\leq h_M(j)h_M(2j),
\qquad 2n\leq2d(M),
$$
which proves <Bernstein inequality for Weyl algebra modules>. The polynomial representation $k[x_1,\ldots,x_n]$ has dimension $n$, so the bound is sharp.
The <associated graded ring> of the <Weyl algebra> is a polynomial ring, so <ascending filtered-graded transfer of Noetherianity> makes $A_n(k)$ a <left Noetherian ring>. Thus every <submodule> of a <finitely generated module> is finitely generated. If $d(M)=n$, every nonzero subquotient has dimension at most $n$ by the exact-sequence formula, and at least $n$ by <Bernstein inequality for Weyl algebra modules>. Every such subquotient therefore has dimension $n$ and positive integral multiplicity.
In a strictly descending chain $M=M_0\supsetneq M_1\supsetneq\cdots$, each nonzero quotient $M_i/M_{i+1}$ consumes at least one unit of multiplicity, by <dimension and multiplicity in a filtered exact sequence>. Hence there are at most $m(M)$ strict steps. This proves the <descending chain condition>, so \b[$M$ is Artinian]. In fact this <holonomic Weyl algebra module> has <module length> at most $m(M)$, as in <multiplicity bounds the length of a holonomic Weyl module>.
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