I want to solve the following :
Independent trials that result in a success with probability $p$ and a failure with probability $1-p$ are called Bernoulli trials. Let $P_n$ denote the probability that n-Bernoulli trials result in an even number of successes (0 being considered an even number). Show that:
$P_n=p(1-P_{n-1})+(1-p)P_{n-1}$ for $n \leq 1$
and use this formula to prove (by induction) that:
$P_n = \frac{1+(1-2p)^{n}}{2}$
My attempt
In the first part I want to say that if the first trial is a success, then the remaining $n-1$ must result in a odd number of successes, whereas If it is a failure then then $n-1$ must result in a even number of successes.
For the second part I have that $P_1 =\frac{1+(1-2p)}{2} = 1-p$, then we assumed that the formula is true for $n-1$ then:
$p(1-\frac{1+(1-2p)^{n-1}}{2}) + (1-p)(\frac{1+(1-2p)^{n-1}}{2})=p-(1+(1-2p)^{n-1})[\frac{p}{2}+\frac{1-p}{2}]=p-\frac{1+(1-2p)^{n-1}}{2}$
Then my questions are Can you help me to accomplish the proof of both parts? please, thanks a lot in advance for your help :)
