<s>[INST] What is the probability that the first head will appear on the even numbered tosses Question

$\text{Consider a coin with probability R to be heads. What is the probability}$
$\text{that the first head will appear on the even numbered tosses?}$

My Approach

let the required Probability$=P$.

Hence we can write our eauation as-:
$$P=(1-R) \times R+(1-R) \times (1-R) \times (1-R) \times P $$
$$P(1-(1-R) \times (1-R) \times (1-R))=(1-R) \times R $$
$$P=\frac{(1-R) \times R }{(1-(1-R) \times (1-R) \times (1-R)}$$
Am i correct?
Answer is given as-:
$$P=\frac{(1 - R)}{(2 - R)}$$
 [/INST] The probability that the first head will appear on the second toss is $(1 - R)R$.  
The probability that the first head will appear on the fourth toss is $(1 - R)^3R$.
The probability that the first head will appear on the sixth toss is $(1 - R)^5R$.  
In general, the probability that the first head will appear on the $2k$th toss is $(1 - R)^{2k - 1}R$.  
Hence, the desired probability is 
$$p = \sum_{k = 1}^{\infty} (1 - R)^{2k - 1}R = (1 - R)R\sum_{k = 1}^{\infty} (1 - R)^{2k}$$
which is a geometric series with common ratio $(1 - R)^2$.  If $R < 1$, we obtain
\begin{align*}
p & = (1 - R)R \cdot \frac{1}{1 - (1 - R)^2}\\
  & = \frac{(1 - R)R}{1 - (1 - 2R + R^2)}\\
  & = \frac{(1 - R)R}{2R - R^2}\\
  & = \frac{(1 - R)R}{R(2 - R)}\\  
  & = \frac{1 - R}{2 - R} && \text{provided that $R \neq 0$}
\end{align*}
If $R = 1$, then heads will be obtained on the first toss, so $p = 0$.  If $R = 0$, then heads will never be obtained, so again $p = 0$.
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