Consider a general chain with the state space $S=\{1,2\}$ and write the transition probability as
$$\begin{pmatrix}
1-a&a\\
b&1-b\end{pmatrix}$$
Use the Markov property to show that
$$P(X\_{n}=1)=\dfrac{b}{a+b}+(1-a-b)^n \left(P(X\_0=1)-\dfrac{b}{a+b}\right)$$.

So, I started off with
\begin{align\*}
P(X\_{n+1}=1)&=P(X\_{n+1}=1|X\_n=1)P(X\_n=1)+P(X\_{n+1}=1|X\_n=2)P(X\_n=2)\\
&=(1-a)P(X\_n=1)+bP(X\_n=2)\\
&=(1-a)P(X\_n=1)+b(1-P(X\_n=1))\\
&=(1-a-b)P(X\_n=1)+b
\end{align\*}
Then, I manipulated the form to show that
\begin{align\*}
P(X\_{n+1}=1)-\dfrac{b}{a+b}&=(1-a-b)P(X\_n=1)-\dfrac{(1-a-b)b}{a+b}\\
P(X\_{n+1}=1)&=(1-a-b)P(X\_n=1)+b
\end{align\*}
Any attempt I made to show that the original statement is true from the result that I got led me nowhere.