So, I figured something out and I'd appreciate it if you guys could try to prove it. $$ F(n+2) = 1 + n + (n-1)C_2 + (n-2)C_3 + (n-3)C_4 +\ldots $$ where $F(n+2)$ is the $(n+2)^{th}$ Fibonacci number.
Here's how I kinda proved it: I got a question in class which was : a fair coin is tossed n times. Let $A(n)$ denote the number of cases in which no two heads occur consecutively and they'd asked different values like $A(5), A(8)$, etc.
Two different approaches to this problem gave me the result I've put.
My teacher took 2 cases-
Last toss is a $T$.
This means that the sequence of $(n-1)$ coins behind it have no two consecutive heads. So basically $A(n-1)$.Last toss is a $H$. This means that the toss just before it has to be $T$. This means that the sequence of $(n-2)$ coins before the $TH$ has no two consecutive heads. So $A(n-2)$.
Using this, we get $A(n) = A(n-1)+A(n-2)$.
$A(1)$ is $2$ and $A(2)$ is $3$. So we basically get the Fibonacci series.
I took several cases and for each case, I used the Gap method. Basically, if there are $x$ times $H$ and $(n-x)$ times $T$, I have $(n-x+1)$ gaps where I can put the $x$ times $H$ = $(n-x+1)C_x$.
So, for $0 H, n T$, we get $1$
$1 H$ and $(n-1) T \ge nC_1$
$2 H$ and $(n-2) T \ge (n-1)C_2$
And so on. $$ A(n) = 1 + n + (n-1)C_2 + (n-2)C_3 + \ldots $$
I was looking for a more valid proof for this. I'd appreciate all the help I can get.
