(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)

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

(4) For $n=1$ the expression gives $2=F_3$

For $n=2$ the expression gives $3=F_4$

For $n=3$ the expression gives $5=F_5$

For $n=4$ the expression gives $8=F_6.$

Conjecture : the result is $F_{n+2},$
We have shown that this conjectureis true for $n=1\dots 4$.
Assume that the result is true for $n=k$ then

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

and hence the result is true for $n=k+1$. By the axiom of induction the result is true for all positive integer values of $n$.