
Binet's formula
Use Binet's formula to show that
$\displaystyle \frac{f_n+1}{f_n} = \frac{1+\sqrt{5}}{2} $. I know that Binet's formula is
$\displaystyle f_n= \frac{(1+ \sqrt{5})^n(1 \sqrt{5})^n}{(2^n)\sqrt{5}} $
So, when I substitute, I get down to
$\displaystyle f_n= \frac{(1+ \sqrt{5})^{n+1}(1 \sqrt{5})^{n+1}}{(1+ \sqrt{5})^{n}(1 \sqrt{5})^n(2^n)\sqrt{5}} $
but I am not sure how to simpify this equation. I know that the answer is
$\displaystyle \frac{1+ \sqrt{5}}{2}$.

That is the limit of $\displaystyle \frac{f_{n+1}}{f_n}$ as n approaches infinity. You can start by expanding the formula algebraically and taking terms common. At the end you will be left with proving that $\displaystyle \frac{1}{ ( \frac{1 + \sqrt{5}}{1  \sqrt{5}} )^n  1 }$ converges to 0 as n approaches infinity.
Notice that you can write $\displaystyle a^{n+1}  b^{n+1}$ as $\displaystyle (ab)( a( a^{n1} + a^{n2}b + .... + b^{n1} ) + b^n )$.