We are given $F_n=\frac{1}{\sqrt{5}}({\alpha }^n-{\beta }^n)$ where $\alpha$ and $\beta$ are solutions of the quadratic equation $x^2-x-1=0$ and $\alpha >\beta$.

(1) In the quadratic equation ${\mathrm{ax}}^2+\mathrm{bx}+c=0$ with roots $\alpha$ amd $\beta$, using Viete's relations for the sum and product of the roots, I obtain: $\alpha \beta =-b/a$, $\alpha +\beta =c/a$
In the particular case of the equation $x^2-x-1=0$, I have: $\alpha \beta =-1$, $\alpha +\beta =1$
(2)

${\displaystyle \frac{1}{\alpha }+\frac{1}{{\alpha }^2}=\frac{\alpha +1}{{\alpha }^2}=1}$

as $\alpha$ satisfies $x^2=x+1$. Similarly for $\beta$.

(3) Here I shall prove that $F_1=1$, $F_2=1$ and $F_n+F_{n+1}=F_{n+2}$ and hence $F_n$ is the $n$th Fibonacci number. First, I calculate $\alpha$ and $\beta$: $\alpha =\frac{1+\sqrt{5}}{2}$ and $\alpha =\frac{1+\sqrt{5}}{2}$

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{F_1}& =\frac{1}{\sqrt{5}}(\frac{1+\sqrt{5}}{2}-\frac{1-\sqrt{5}}{2})=1\\ \multicolumn{1}{c}{F_2}& =\frac{1}{\sqrt{5}}({\alpha }^2-{\beta }^2)=\frac{1}{\sqrt{5}}(\alpha -\beta )(\alpha +\beta )=1\\ \multicolumn{1}{c}{F_n+F_{n+1}-F_{n+2}}& =\frac{1}{\sqrt{5}}[{\alpha }^n-{\beta }^n+{\alpha }^{n+1}-{\beta }^{n+1}-{\alpha }^{n+2}+{\beta }^{n+2}]\\ \multicolumn{1}{c}{F_n+F_{n+1}-F_{n+2}}& =\frac{1}{\sqrt{5}}[-{\alpha }^n({\alpha }^2-\alpha -1)+{\beta }^n({\beta }^2-\beta -1)]=0\end{array}}$

(4) I shall prove by induction the statement $P_n$ that $1+F_1+F_2+\dots F_n=F_{n+2}$.

I know that $F_1=1,F_2=1$ and by (3), $F_3=2,F_4=3,F_5=5,F_6=8,...$
For $n=1$, $P_1$ is $1+F_1=F_3$ which is evidently true as 1 + 1 = 2.
For $n=2$, $P_1$ is $1+F_1+F_2=F_4$ which is evidently true as 1 + 1 + 1 = 3.
For $n=3$, $P_1$ is $1+F_1+F_2+F_3=F_5$ which is evidently true as 1 + 1 + 1 + 2 = 5.

Now I assume that $P(k)$ is true for a fixed $k$ and I shall prove that $P(k+1)$ is also true, that is:

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{P_k}& :1+F_1+F_2+\dots F_k=F_{k+2}\\ \multicolumn{1}{c}{P_{k+1}}& :1+F_1+F_2+\dots F_k+F_{k+1}=F_{k+3}\end{array}}$

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{1+F_1+F_2+\dots F_k+F_{k+1}}& =(1+F_1+F_2+\dots F_k)+F_{k+1}\\ \multicolumn{1}{c}{}& =F_{k+2}+F_{k+1}\\ \multicolumn{1}{c}{}& =F_{k+3}\end{array}}$

and hence the result is true for $n=k+1$. By the principle of mathematical induction the statement $P_n$ is true for all positive integer values of $n$ which completes the proof.