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Given $F_n={1\over\sqrt5}(\alpha^n-\beta^n)$ where $\alpha$ and $\beta$ are solutions of the quadratic equation $x^2-x-1=0$ and $\alpha > \beta$ show that

(1)$\alpha\beta =-1$, $\alpha + \beta = 1$

(2)${ 1\over \alpha}+{1\over \alpha^2} = {1\over \beta} + {1\over \beta^2}=1$

(3)$F_1=F_2=1$ and $F_n + F_{n+1} = F_{n+2}$ and hence $F_n$ is the $n$th Fibonacci number and

(4) the sum $1 + F_1 + F_2 + \ldots F_n$ gives another Fibonacci number.