= Solution
For the <sawtooth Ising chain>, each apex <spin> occurs in only one triangle. Put $K=\beta J$. Summing that <spin> first gives the effective <transfer matrix>
$$
T_{ss'}=\sum_{t=\pm1}\exp\{K[ss'+t(s+s')]\}
=2e^{Kss'}\cosh[K(s+s')].
$$
In the ordered basis $s=+1,-1$ this is
$$
T=\begin{pmatrix}2e^K\cosh2K&2e^{-K}\\2e^{-K}&2e^K\cosh2K\end{pmatrix}.
$$
Its symmetric and antisymmetric <eigenvectors> are $(1,1)/\sqrt2$ and $(1,-1)/\sqrt2$, with <eigenvalues>
$$
\lambda_+=e^{3K}+3e^{-K},\qquad\lambda_-=e^{3K}-e^{-K}.
$$
Consequently
$$
\boxed{Z_L=(e^{3K}+3e^{-K})^L+(e^{3K}-e^{-K})^L,\qquad
f_\infty=-\frac{k_BT}{2}\log(e^{3K}+3e^{-K}).}
$$
At $J=0$, the result is $Z_L=4^L$, correctly counting $2L$ independent <Ising spins>. As $T\to0$, $f_\infty\to-3J/2$, the energy of three satisfied ferromagnetic bonds per two sites.
The printed summand lacks grouping brackets. The result above uses the translation-invariant common-$J$ interpretation. If base and apex bonds instead have independent couplings $J_b,J_a$, the same derivation gives $\lambda_\pm=2e^{\beta J_b}\cosh(2\beta J_a)\pm2e^{-\beta J_b}$; specifying those couplings is essential before giving a different answer.
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