Probability of an even number of successes in a Binomial distribution

We define

1. $X\sim \mathcal{B}in(n,p)$ to be a Random variable following a Binomial Distribution.
2. $a_n:=\mathbb{P}[X{\equiv }_{2}0]$, i.e. the probability of $X$ being even.

Then $a_{n+1}=p(1-a_n)+(1-p)a_n=p+(1-2p)a_n$. This is a Linear Difference Equation, which has general solution $a_n=\alpha +\beta (1-2p{)}^n$. Since $p_0=1$, we get $\alpha +\beta =1$. Matching coefficients we get:

$$\begin{array}{rlc}{a}_{n+1}& =p+(1-2p)a_n& \\ \alpha +\beta (1-2p{)}^{n+1}& =p+(1-2p)(\alpha +\beta (1-2p{)}^n)& \\ \alpha +\beta (1-2p{)}^{n+1}& =p+\alpha -2p\alpha +\beta (1-2p{)}^{n+1}& \\ \alpha & =p+\alpha -2p\alpha & \\ 0& =p-2p\alpha & p\ne 0\\ 0& =1-2\alpha & \\ \alpha & =\frac{1}{2}& \end{array}$$

Hence $\mathbb{P}[X{\equiv }_{2}0]=\frac{1+(1-2p{)}^n}{2}$