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Adding the ordinary differential equations for all enzyme states makes the net rates telescope:With initially unbound enzyme, this sum is $E_0$ for all time. HenceThis is enzyme conservation, not conservation of each complex individually. The reaction changes site occupancy and makes reaction product, but no event creates or destroys an enzyme molecule.

For an outer solution with bounded derivatives as $ϵ\to 0$, the final complex equation gives $r_n=0$. Working backwards in $0=r_i-r_{i+1}$ then gives every $r_i=0$, including $r_1$; setting only their sum to zero would not suffice. This is the quasi-steady-state approximation, valid after the initial complex-formation layer with the relevant dimensionless rate parameters held fixed.

The recurrence is $C_i=S{C}_{i-1}/K_i$, where $K_i=(k_{-i}+k_i^{+})/k_i$. Thus $C_j={\varphi }_j{S}^j{C}_0$, ${\varphi }_j={\prod }_{i=1}^j{K}_i^{-1}$. Substitution into enzyme conservation yieldsEach state $j$ produces reaction product at rate $k_j^{+}{C}_j$, soThe source's uppercase $N$ in the index range is read as the number $n$ of binding sites. This approximation preserves the total enzyme population and the different catalytic rates of the occupied states.

For sequential enzyme binding, a quasi-steady-state approximation gives occupied-state weights ${\varphi }_j{S}^j$, ${\varphi }_j={\prod }_{i=1}^j{K}_i^{-1}$, relative to free enzyme. If all intermediate weights tend to zero while the terminal weight remains nonzero, enzyme conservation leaves only unbound and fully bound states, and the production rate becomes the displayed Hill equation. One realization holds other rates fixed while the first binding rate tends to zero and the last tends to infinity with positive finite product. A statement merely bounding that product does not by itself specify this nontrivial distinguished limit.