Prove that every pair of consecutive Fibonacci numbers are coprime. That is, they share no common factors other than 1.
Calculate the following: \(F_1^2-F_0F_2\), \(F_2^2-F_1F_3\), \(F_3^2-F_2F_4\), \(F_4^2-F_3F_5\) and \(F_5^2-F_4F_6\). What do you notice?
Work out \(F_3^2-F_0F_6\), \(F_4^2-F_1F_7\), \(F_5^2-F_2F_8\) and \(F_6^2-F_3F_9\). 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 \(\sum_{i=0}^nF_i^2=F_0^2+F_1^2+F_2^2+...+F_{n-1}^2+F_n^2\) 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?
In the example, we saw that \(\varphi^2=\varphi+1\). Can you write \(\varphi^3\) in the form \(a\varphi+b\), where \(a\) and \(b\) are positive integers?
Simplify \(F_0-F_1+F_2-F_3+...-F_{2n-1}+F_{2n}\), where \(n\) is a positive integer.
In the \(6\times7\) large rectangle shown below, how many rectangles are there in total formed by grid lines? [need to insert image]
How many cuboids are contained in an \(n\times n\times n\) cube? For example, we’ve got \(n^3\) cuboids of size \(1\times1\times1\), and obviously just \(1\) of size \(n\times n\times n\) (which is the whole cube itself). But we also have to count how many there of size \(1\times1\times2\), \(1\times2\times3\), and several more.