A coin has probability $p$ of showing head when tossed. It is tossed $n$ times. Let $P_n$ be the probability that no two or more consecutive heads occur? What is $P_n$ for any $n≥3$. I proved it individually fo
No heads together?
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2[This guide](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) contains a lot of pointers for how to format mathematical text. Specifically, `$P_n$` gives $P_n$. – 2017-02-20
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Let $u_n$ be the number of sequences of length $n$ composed with T and H such that there are no two consecutive H. We have $u_1=2$ and $u_2=3$.
A sequence of $n+1$ throws can end with a T, there are $u_n$ such sequences, or by a H, which has to be preceded by a T. There are $u_{n-1}$ such sequences.
So $(u_n)$ follow the recurrence : $$u_1=2,\,u_2=3\text{ and }(\forall n\ge2)\,u_{n+1}=u_n+u_{n-1}$$ So $(u_n)$ is simply a "shifted" Fibonacci sequence.
With your usual method to solve those recurrences, you find : $$u_n=\frac{5+3\sqrt5}{10}(\frac{1+\sqrt5}{2})^n + \frac{5-3\sqrt5}{10}(\frac{1-\sqrt5}{2})^n$$ And the seeked probability is $$P_n=\frac{u_n}{2^n}$$