An nth term for a Fibonacci series? Say the first two terms of a sequence are $a_0,a_1$, then the remaining terms meet the formula $$a_{n+2}=a_{n+1}+a_n$$
What is the $n_{th}$ term formula?
I figured that $$\lim_{n \rightarrow \infty}\frac{a_{n+1}}{a_n}=\Phi = \frac{1+\sqrt 5}{2}\approx 1.618$$
Using this fact, find the $n_{th}$ term formula for the Fibonacci Series. 

Let $2 \alpha = 1 + \sqrt{5}$ and $2 \beta = 1 - \sqrt{5}$, where $\alpha, \beta$ are roots of $x^2 = x + 1$, then given $a_{n+2} = a_{n+1} + a_{n}$ it is seen that:
$$a_{n} = A \, \alpha^n + B \, \beta^n. $$
Now
\begin{align}
a_{0} &= A + B \\
a_{1} &= A \, \alpha + B \, \beta
\end{align}
which yields $(\alpha - \beta) \, A = a_{1} - a_{0} \beta$, $(\alpha - \beta) \, B = a_{0} \alpha - a_{1}$, and
$$a_{n} = \frac{1}{\alpha - \beta} \, (a_{1} \, (\alpha^n - \beta^n) + a_{0} \, (\alpha^{n-1} - \beta^{n-1})) = a_{1} \, F_{n} + a_{0} \, F_{n-1}. $$
Here 
$$F_{n} = \frac{\alpha^n - \beta^n}{\alpha - \beta}$$
are the Fibonacci numbers. 
Given: $(a_{0}, a_{1}) = (0,1)$ then $a_{n} = F_{n}$ and $(a_{0}, a_{1}) = (2,1)$ then $a_{n} = F_{n} + 2 \, F_{n-1} = F_{n+1} + F_{n-1} = L_{n}$, where $L_{n}$ are the Lucas numbers. 
