Solution
= Solution
On the class $\{1,2\}$, the two-step <transition matrix> is symmetric with positive entries, so its <stationary distribution> is $(1/2,1/2)$ and its distribution converges to that law. Therefore
$$
\boxed{\lim_{n\to\infty}P(X_{2n}=2\mid X_0=1)=\frac12.}
$$
Explicitly the two eigenvalues are $1$ and $1/9$, giving $P(X_{2n}=2\mid X_0=1)=\tfrac12(1-9^{-n})$. The original chain has probability zero of being in state $2$ at odd times, consistent with its period two.