= Solution
Let $M=L/2$, so that the total number of <Ising spins> is $3M$. Write $K=\beta J$ and group each triangle's two base spins and its apex into one cell. The <Ising chain with alternating triangular decorations> has a triangle bond followed by an undecorated connecting bond. Summing an apex <Ising spin> first gives a decorated-bond <Boltzmann weight>
$$
A_{st}=\sum_{\tau=\pm1}e^{-K[st+(s+t)\tau]}=2e^{-Kst}\cosh[K(s+t)].
$$
The undecorated bridge has <Boltzmann weight> $B_{ts'}=e^{-Kts'}$. In the basis $+1,-1$,
$$
A=\begin{pmatrix}e^K+e^{-3K}&2e^K\\2e^K&e^K+e^{-3K}\end{pmatrix},\qquad
B=\begin{pmatrix}e^{-K}&e^K\\e^K&e^{-K}\end{pmatrix}.
$$
Each cell contributes $T=AB$, so <periodic boundary conditions> give
$$
\boxed{Z_L=\operatorname{Tr}(AB)^M.}
$$
The two <matrices> have common normalized <eigenvectors> $(1,1)^{\mathsf T}/\sqrt2$ and $(1,-1)^{\mathsf T}/\sqrt2$, expressing <spin inversion symmetry>. Their respective <eigenvalues> are
$$
a_+=3e^K+e^{-3K},\quad a_-=e^{-3K}-e^K,\qquad
b_+=e^{-K}+e^K,\quad b_-=e^{-K}-e^K.
$$
Consequently the cell <transfer matrix> has <eigenvalues>
$$
\begin{aligned}
\lambda_+&=(3e^K+e^{-3K})(e^{-K}+e^K)=3+e^{-4K}+e^{-2K}+3e^{2K},\\
\lambda_-&=(e^{-3K}-e^K)(e^{-K}-e^K)=e^{-4K}-e^{-2K}-1+e^{2K},
\end{aligned}
$$
and the finite-size <partition function> is
$$
\boxed{Z_L=\lambda_+^{L/2}+\lambda_-^{L/2}.}
$$
For $K>0$, $\lambda_-\geq0$ and $\lambda_+-\lambda_-=4+2e^{-2K}+2e^{2K}>0$, so only $\lambda_+$ survives in the <thermodynamic limit>. Divide the <Helmholtz free energy> by $3M$, the number of sites rather than the number of cells or base spins:
$$
\boxed{f(T)=-\frac{k_BT}{3}\ln\lambda_+
=-\frac{k_BT}{3}\ln\!\left(3+e^{-4\beta J}+e^{-2\beta J}+3e^{2\beta J}\right).}
$$
As a counting check, $K=0$ gives $\lambda_+=8$, $\lambda_-=0$, hence $Z_L=8^M=2^{3M}$, exactly the number of independent <Ising spin> configurations. This model has separated decorated bonds and should not be confused with the corner-sharing <sawtooth Ising chain>, which has a different site count.
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