Note that this is a simplified version of a problem from a stochastic processes course.
Given transition matrix $P$, show that $p_{1,1}^{(n)} = .25 + .75(1 - 4\alpha)^n$ where
$$P = \begin{bmatrix} 1 - 3\alpha & \alpha & \alpha & \alpha\\ \alpha & 1 - 3\alpha & \alpha & \alpha\\ \alpha & \alpha & 1 - 3\alpha & \alpha\\ \alpha & \alpha & \alpha & 1 - 3\alpha \end{bmatrix}$$
and $\alpha = [0,1/3]$. That is, the probability of changing from one state to another is $3\alpha$ and all other states have equally likely probability of being transitioned to.
So here is where I've gotten so far. I decided that induction is likely the best approach to solve this, and showing that $p_{1,1}^{(0)}$ is trivial, similarly $$p_{1,1}^{(1)} = 1 - 3\alpha\\ = .25 + .75 - 3\alpha \\= .25 + .75(1 - 4\alpha)^1$$
So the trivial case is proven.
Now, assuming $$p_{1,1}^{(k)} = .25 + .75(1 - 4\alpha)^k$$ I must show that $$p_{1,1}^{(k+1)} = .25 + .75(1 - 4\alpha)^{k+1}$$ My thought process here is that we sum the total probabilities. $$p_{1,1}^{(k+1)} = p_{1,1}^{(k)}p_{1,1}^{(1)} + p_{1,2}^{(k)}p_{2,1}^{(1)} + p_{1,3}^{(k)}p_{3,1}^{(1)} + p_{1,4}^{(k)}p_{4,1}^{(1)}\\ =[.25 + .75(1-4\alpha)^k](1 - 3\alpha) + \alpha(p_{1,2}^{(k)}+p_{1,3}^{(k)}+p_{1,4}^{(k)})$$ But I'm not sure where to go from here.
This is thanks to the comments of saulzpatz, but by reducing $p_{1,2}^{(k)} + p_{1,3}^{(k)} + p_{1,4}^{(k)}$ to $1 - p_{1,1}^{(k)}$ by properties of a stochastic matrix, the algebra reduces quite beautifully.
$$\alpha(1 - p_{1,1}^{(k)}) + (1 - 3\alpha)p_{1,1}^{(k)}\\ =\alpha + p_{1,1}^{(k)} - 4\alpha p_{1,1}^{(k)}\\ =\alpha + p_{1,1}^{(k)}(1-4\alpha)\\ =\alpha + .25(1-4\alpha) + .75(1-4\alpha)^k(1-4\alpha)\\ =.25 + .75(1-4\alpha)^{k+1} = p_{1,1}^{k+1}$$