Fibonacci function Dear Professors and Mathematcians,
Now, I am introducing Fibonacci sequence and function.
Consider, $F(x)$ is a Fibonacci function and $f_n$ is Fibonacci sequence. For fixing the initial values by definition, like $f_0= 0$, $f_1=1=f_2$. I made the following true proposition which is true by trial and error method for $n\geq2$ and $n$ is some integer. For all real value of $x$, the following proposition is true. But, we can state as a theorem if one can produce a proof. Of course, I failed to prove and seeking helps to prove the statement.
$$F(x +n) = f_nF(x+1)+f_{n-1}F(x).$$
Thanks in advance
 A: I assume that you define the Fibonacci function as the natural continuation of the Fibonacci sequence,
$$
F(x) = \frac{\alpha^x-\beta^x}{\sqrt5}
$$
where $\alpha = (1+\sqrt5)/2, \beta=(1-\sqrt5)/2$. If not, I'll just delete this answer and will have wasted my time.
For later use, note that $\alpha\beta+1=0,\alpha+1/\alpha=\sqrt5\text{, and }\beta+1/\beta=-\sqrt5.$
Now notice that for any integer $n$ we have $F(n)=f_n,$ so you wish to prove
$$
F(x+n)=F(n)F(x+1)+F(n-1)F(x)
$$
Using the supposed definition of $F(x)$ the right side of your equality, $F(n)F(x+1)+F(n-1)F(x)$, becomes 
$$
\begin{align*}
&= \frac{1}{5}[(\alpha^n-\beta^n)(\alpha^{x+1}-\beta^{x+1})+(\alpha^{n-1}-\beta^{n-1})(\alpha^x-\beta^x)]\\
&= \frac{1}{5}[\alpha^{x+n+1}-\alpha^n\beta^{x+1}-\alpha^{x+1}\beta^n+\beta^{x+n+1}+\alpha^{x+n-1}-\alpha^{n-1}\beta^x-\alpha^x\beta^{n-1}+\beta^{x+n-1}]\\
&=\frac{1}{5}[\alpha^{x+n}(\alpha+1/\alpha)+\beta^{x+n}(\beta+1/\beta)-\alpha^{n-1}\beta^x(\alpha\beta+1)-\alpha^x\beta^{n-1}(\alpha\beta+1)]\\
&=\frac{1}{5}[\alpha^{x+n}(\alpha+1/\alpha)+\beta^{x+n}(\beta+1/\beta)]\quad\text{(using our first observation above)}\\
&=\frac{1}{5}[\alpha^{x+n}\sqrt5-\beta^{x+n}\sqrt5]\quad\text{(using our second and third observations)}\\
&=\frac{\alpha^{x+n}-\beta^{x+n}}{\sqrt5}\\
&=F(x+n)
\end{align*}
$$
A: It also can be proven by induction on $n$ :


*

*for $n = 1$, we have: 
$$
F(x + 1) = f_1 F(x + 1) + f_0 F(x) = F(x + 1)
$$

*assuming the equality holds for $\forall i \leq n$, we'll prove that it's true for $n + 1$. Using the equality $F(x + 2) = F(x) + F(x + 1)$, it follows:
$$
\begin{align*}
&F(x+n+1) = F(x+n) + F(x+n-1)\\
&= f_n F(x+1) + f_{n-1} F(x) + f_{n-1} F(x+1) + f_{n-2} F(x)\\
&= (f_n + f_{n-1}) F(x+1) + (f_{n-1} + f_{n-2}) F(x)\\
&= f_{n+1} F(x+1) + f_n F(x)\\
\end{align*}
$$

