General Two-State Markov Chain: $P(X_{n}=1)=\frac{b}{a+b}+(1-a-b)^n \big(P(X_0=1)-\frac{b}{a+b}\big)$ 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.
 A: From 
$$P(X_{n+1}=1)=(1-a-b)P(X_n=1)+b,$$
You can conclude that
\begin{align}
P(X_{n+1}=1)&=(1-a-b)\left[(1-a-b)P(X_{n-1}=1)+b\right]+b\\
&=(1-a-b)^2P(X_{n-1}=1)+(1-a-b)b+b
\end{align}
Repeating the procedure, we have
\begin{align}
P(X_{n}=1)&=(1-a-b)^nP(X_0=1)+b \sum_{i=0}^{n-1}(1-a-b)^i\\
&=(1-a-b)^nP(X_0=1)+b \left[\frac{1-(1-a-b)^n}{1-(1-a-b)}\right]\\
&=(1-a-b)^nP(X_0=1)+b \left[\frac{1-(1-a-b)^n}{a+b}\right]\\
&=\frac{b}{a+b}+(1-a-b)^n\left[P(X_0=1)-\frac{b}{a+b}\right]
\end{align}
A: Hint. From the relation
$$
u_{n+1}=(1-a-b)u_n+b,\quad n\geq0, \tag1
$$ let's find a fixed real number $\alpha$ such that
$$
\left(u_{n+1}-\alpha\right)=(1-a-b)\left(u_n-\alpha\right). \tag2
$$ 
Inserting $(1)$ in $(2)$ and expanding the right hand side gives
$$
\color{red}{(1-a-b)u_n}+b-\alpha=\color{red}{(1-a-b)u_n}-\alpha(1-a-b)
$$ that is, after symplyfying,

$$
\alpha=\frac{b}{a+b}, \qquad a+b \neq 0.\tag3
$$ 

Now from $(2)$, we get that as $u_n-\alpha$ is a geometric sequence: 

$$
u_n-\alpha=(1-a-b)^n\left(u_0-\alpha\right) \tag4
$$ 

equivalently, using $(3)$,

$$
u_n=\dfrac{b}{a+b}+(1-a-b)^n \left(u_0-\dfrac{b}{a+b}\right)\tag5
$$ 

as wanted.
