Fibonacci sequence

Prove the following statements about the Fibonacci sequence \(F_1=F_2=1\), \(F_n=F_{n-1}+F_{n-2}\):

Variant 1
\(\displaystyle F_n\le \left(\frac{1+\sqrt{5}}{2}\right)^{n-1} \).
Resolution
For \(n=1\) we have \(1\le 1\).

For \(n=2\) we have \(1\le \frac{1+\sqrt 5}2\).

We show that the statement is true for \(n+1\) if it holds for \(n\) and \(n-1\):
\(\displaystyle F_{n+1}=F_n+F_{n-1}\le \left(\frac{1+\sqrt{5}}{2}\right)^{n-1}+\left(\frac{1+\sqrt{5}}{2}\right)^{n-2}= \left(\frac{1+\sqrt{5}}{2}\right)^{n-2}\left(\frac{3+\sqrt{5}}{2}\right)=\\ \left(\frac{1+\sqrt{5}}{2}\right)^{n-2}\left(\frac{1+\sqrt{5}}{2}\right)^2= \left(\frac{1+\sqrt{5}}{2}\right)^{n}\)
Variant 2
\( \displaystyle \sum_{i=1}^{n}F_i =F_{n+2}-1 \).
Resolution
For \(n=1\) we have \(1=2-1\).

We show that the statement is true for \(n+1\) if it holds for \(n\):
\( \displaystyle \sum_{i=1}^{n+1}F_i=\sum_{i=1}^{n}F_i+F_{n+1}=F_{n+2}-1+F_{n+1}=F_{n+3}-1 \)
Variant 3
\( \displaystyle \sum_{i=1}^{n}F_i^2 =F_nF_{n+1} \).
Resolution
For \(n=1\) we have \(1^2=1{\cdot}1\).

We show the statement is true for \(n+1\) if it holds for \(n\):
\( \displaystyle \sum_{i=1}^{n+1}F_i^2= \sum_{i=1}^{n}F_i^2+F_{n+1}^2= F_nF_{n+1}+F_{n+1}^2= F_{n+1}(F_n+F_{n+1})= F_{n+1}F_{n+2} \)
Variant 4
\( \displaystyle \sum_{i=1}^{n}F_{2i-1} =F_{2n} \).
Resolution
For \(n=1\) we have \(1=1\).

We show that the statement is true for \(n+1\) if it holds for \(n\):
\( \displaystyle \sum_{i=1}^{n+1}F_{2i-1}= \sum_{i=1}^{n}F_{2i-1}+F_{2n+1}= F_{2n}+F_{2n+1}= F_{2(n+1)} \)