Why the $GCD$ of any two consecutive Fibonacci numbers is $1$? Note: I've noticed that this answer was given in another question, but I merely want to know if the way I'm using could also give me a proof.
I did the following:
$$F_n=F_{n-1}+F_{n-2} \\ F_n=[F_{n-2}+F_{n-3}]+[F_{n-3}+F_{n-4}]\\F_n=F_{n-2}+2F_{n-3}+F_{n-4}\\ F_n=[F_{n-3}+F_{n-4}]+2[F_{n-4}+F_{n-5}]+[F_{n-5}+F_{n-6}]\\F_n=F_{n-3}+3F_{n-4}+3F_{n-5}+F_{n-6}\\ \dots \tag{1}$$
I guess that the coefficients of the $F_n$'s might indicate something that could prove it. But I'm not sure if it's possible. Perhaps the impossibility of writing the expression as:
$$n(b_1a_1+b_2a_2+\dots +b_n a_n)$$
With $n,b_n\in\mathbb{N}$ would show that. But I'm not sure on how to proceed. This should be true because if $a$ and $b$ have a common divisor $d$, then:
$$a+b=a'd+b'd=d(a'+b')$$
It is possible to extend this:
$$a+b+c=da'+db'+dc'=d(a'+b'+c')$$
I have noticed that the numbers that appear in the expansion I've shown in $(1)$ seems to be the Pascal's triangle. So perhaps these numbers as coefficients of the $F_n$'s might indicate that it's not possible to write them as: 
$$d(a'+b'+c')$$

Let us take $3$ consecutive terms, call them $a, b, a+b$, with $a<b$.
As $\gcd(b,a+b)=\gcd(a,b)$, we can by iterating see that the gcd is the same for any pair of consecutive Fibonacci numbers and we can conveniently calculate it for the first two numbers which is $1$.
