Biased coin equation when odd number of head occured Let $C$ be a biased coin such that the probability of a head turning up is $p.$ Let $p_n$ denote the probability that an odd number of heads occurs after $n$ tosses for $n \in \{0,1,2,\ldots \},$ Then which of the following is TRUE ?
$A)p_n=\dfrac{1}{2}\text{ for all }n \in \{0,1,2,\ldots \}.$
$B)p_n=(1-p)(1-p_{n-1})+p.p_{n-1}\text{ for }n\geq 1\text{ and }p_{0}=0.$
$C)p_{n}=\sum_ {i=1}^{n}p(1-2p)^{i-1} \text{ for }n\geq 1.$
$D)\text{ If } p=\dfrac{1}{2},\text{ then } p_{n}=\dfrac{1}{2} \text{ for all }n \in \{0,1,2,\ldots\}$.
$E)p_{n}=1 \text{ if } n \text{ is odd and } 0 \text{ otherwise}.$

My approach
$P_{0}=0$
 $P_{1}={H}=1/2$
$P_{2}={HT,TH}=2/4=1/2$
$P_{1}={THT,TTH,HTT,HHH}=4/8=1/2$
Now, check option B)
Here, Say for $n=3$
$(1-1/2)(1-(1/2.1/2))+1/2.(1/2.1/2)=1/2$
Is it correct?
 A: For any any n:
Probability of getting odd number of heads is nothing but the sum of odd terms of a binomial distribution 
$(p+1-p)^n = {n\choose0}p^0(1-p)^n+{n\choose1}p(1-p)^{n-1}+{n\choose2}p^2(1-p)^{n-2}+\cdots +{n\choose n}p^n(1-p)^0 = 1\tag1$
$(1-p-p)^n = {n\choose0}p^0(1-p)^n-{n\choose1}p(1-p)^{n-1}+{n\choose2}p^2(1-p)^{n-2}+\cdots -{n\choose n}p^n(1-p)^0 = (1-2p)^n\tag2$
Subtract (2) from (1), you get terms with powers of p to be odd to have odd number of heads.
Sum of odd numbers $p_n= \frac{1}{2}[(p+1-p)^{n} - (1-2p)^{n}]$
$p_n=\frac{1}{2}\left(1-(1-2p)^n\right)$
Choice C $\sum_{i=1}^{n-1} p(1-2p)^i = p.\dfrac{1-(1-2p)^n}{2p} = \frac{1}{2}\left(1-(1-2p)^n\right)$
Reduces to the above expression and hence the answer is (C)
A: Proof by counterexample/elimination.
D cannot be true, as $p_0 = 0$.
Assume $ p = 1$. Then all coins are going to turn up heads. $p_n$ is 1 for odd $n$ and $0$ for even $n$.
A is now obviously false in this particular case; therefore, it cannot be a true statement in general.
Calculating $p_1$ using $B$ gives
$$ p_1 = (1 - 1)(1 - 0) + 1 \cdot 0 = 0$$
but $p_1 = 1$. So B is false.
Assume $p = 0$. Then all coins are going to turn up tails; the number of heads is always zero, so $p_n = 0$. 
E is therefore false.
We conclude that C must be true, as all other statements are false.
