= Solution
For the <hydraulic ladder network>, using the frontier star just defined, start at a single groove with $(x_1,y_1,z_1)=(0,0,R_g)$. Put $B=y_n+R_c$, $C=z_n+R_c$ and $S=B+C+R_g$. The exact update is
$$
\boxed{x_{n+1}=x_n+\frac{BC}{S},\qquad y_{n+1}=\frac{BR_g}{S},\qquad z_{n+1}=\frac{CR_g}{S}.}
$$
After all grooves have been included, the left frontier is a blind end, so no external flow passes through $y_N$. The inlet-to-outlet resistance is \b[$R_N=x_N+z_N$]. Iterate 49 times to obtain $R_{50}$.
An equivalent closed form, for $R_c,R_g>0$, is
$$
\boxed{R_N=\frac{(N-1)R_c}{2}+\frac{R_c}{\alpha+\sinh\kappa\,\tanh[(N-1)\kappa/2]},\quad \alpha=\frac{R_c}{R_g},\quad\cosh\kappa=1+\alpha.}
$$
Thus substitute $N=50$; no single number follows without $R_c/R_g$. Checks are $R_1=R_g$, $R_2=(R_g+R_c)/2$, and $R_N\to R_g/N$ as $R_c\to0$. The formula counts only the $N-1$ inter-groove segments in each rail; extra inlet/outlet lead resistances, if present, must be added separately. All elements obey linear laminar-flow pressure-drop laws.
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