= Solution
In the $s=+1,-1$ basis the <transfer matrix> and its <eigenvalues> are
$$
A=\begin{pmatrix}a&b\\b&a\end{pmatrix},\qquad a=2e^{-\beta J}\cosh(2\beta\bar J),\quad b=2e^{\beta J},\qquad \lambda_\pm=a\pm b.
$$
For finite positive <temperature>, $a,b>0$ and $\lambda_+>|\lambda_-|$. Consequently $Z=\lambda_+^L+\lambda_-^L$ and $L^{-1}\log Z\to\log\lambda_+$ in the <thermodynamic limit>. There are $2L$ <spins>, so the <Helmholtz free energy> per <spin> is
$$
\boxed{f(\beta,J,\bar J)=-\frac1{2\beta}\log\left[2e^{-\beta J}\cosh(2\beta\bar J)+2e^{\beta J}\right].}
$$
The factor $1/2$ is essential: one <transfer matrix> step counts two <spins>. For completeness, if the undefined $r$ is retained, set $a_r=A^{(r)}_{-1,-1}$, $d_r=A^{(r)}_{r+1,r+1}$ and $b_r=A^{(r)}_{-1,r+1}$. The same <thermodynamic limit> gives $f_r=-(2\beta)^{-1}\log[(a_r+d_r+\sqrt{(a_r-d_r)^2+4b_r^2})/2]$ whenever these are two distinct <spin> values.
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