= Solution
Absorb $\beta$ into the nonconstant couplings and write the corresponding dimensionless action as $H_u$. Choose its additive-constant convention so that one blocking step has
$$
\sum_\sigma K(\sigma'\mid\sigma)e^{-H_u(\sigma)}=e^{-Ng(u)}e^{-H_{Ru}(\sigma')}.
$$
The function $g(u)$ is the <identity-operator contribution to renormalization-group free energy>: it records the extensive constant generated by eliminating short-distance degrees of freedom. Summing over $\sigma'$ gives
$$
Z(u,0,N)=e^{-Ng(u)}Z(Ru,0,N'),\qquad N'=b^{-D}N.
$$
For the dimensionless <free-energy density> $\Phi(u)=-N^{-1}\ln Z(u,0,N)$, this is the inhomogeneous recursion
$$
\boxed{\Phi(u)=g(u)+b^{-D}\Phi(Ru).}
$$
For the explicit constant $C$ in the original <partition function>, the corresponding new constant obeys $\beta C'=b^D[\beta C+g(u)]$; $C$ and $g$ are not additional interactions between spins.
Iterating the recursion $n$ times yields
$$
\Phi(u)=\sum_{j=0}^{n-1}b^{-jD}g(R^ju)+b^{-nD}\Phi(R^nu).
$$
An analytic background $a(u)$ satisfying $a(u)=g(u)+b^{-D}a(Ru)$ absorbs the regular contribution. The remaining singular part $f_s=\Phi-a$ obeys
$$
f_s(u)=b^{-D}f_s(Ru).
$$
This separation explains both the scaling of the singular part and the role of the inhomogeneous term; simply dropping the generated constant would incorrectly change the full <free energy>.
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