Problems

Simplify $F_0-F_1+F_2-F_3+...-F_{2n-1}+F_{2n}$, where $n$ is a positive integer.

Prove that every pair of consecutive Fibonacci numbers are coprime. That is, they share no common factors other than 1.

Calculate the following: $F_1^2-F_0{F}_2$, $F_2^2-F_1{F}_3$, $F_3^2-F_2{F}_4$, $F_4^2-F_3{F}_5$ and $F_5^2-F_4{F}_6$. What do you notice?

Work out $F_3^2-F_0{F}_6$, $F_4^2-F_1{F}_7$, $F_5^2-F_2{F}_8$ and $F_6^2-F_3{F}_9$. What pattern do you spot?

Can every whole number be written as the sum of two Fibonacci numbers? If yes, then prove it. If not, then give an example of a number that can’t be. The two Fibonacci numbers don’t have to be different.

What’s $\sum _{i=0}^n{F}_i^2=F_0^2+F_1^2+F_2^2+...+F_{n-1}^2+F_n^2$ in terms of just $F_n$ and $F_{n+1}$?

What are the ratios $\frac{F_2}{F_1}$, $\frac{F_3}{F_2}$, and so on until $\frac{F_7}{F_6}$? What do you notice about them?

$\phi =\frac{1+\sqrt{5}}{2}$ is the golden ratio. Using the fact that ${\phi }^2=\phi +1$, can you express ${\phi }^3$ in the form $a\phi +b$, where $a$ and $b$ are positive integers?