# Binet's formula

• Sep 28th 2010, 12:06 PM
tarheelborn
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
$\displaystyle \frac{1+ \sqrt{5}}{2}$.
• Sep 28th 2010, 01:13 PM
Traveller
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 $\displaystyle (a-b)( a( a^{n-1} + a^{n-2}b + .... + b^{n-1} ) + b^n )$.