A Non-recursive Fibonacci Sequence How can I determine the general term of a Fibonacci Sequence? Is it possible for any one to calculate F2013 and large numbers like this?
Is there a general formula for the nth term of the Fibonacci Sequence?
If there is a combinatorial formula, that is more preferable; some formula like 
Fn = xCy or xPy or something like that for some x and y. 

When
(1) $F_{n+2} = F_{n+1} + F_{n}$
one can try $F_n = \phi^n$, then we obtain
(2) $\phi^{n+2} - \phi^{n+1} - \phi^n = 0$,
thus
(3) $\phi^n \Big( \phi^2 - \phi - 1 \Big) = 0$,
ignoring the trivial case $F_n = 0$ we obtain
(4) $\phi = \frac{1}{2} \pm \frac{1}{2} \sqrt{5}$
So we get
(5) $F_n = a \left( \frac{1}{2} + \frac{1}{2} \sqrt{5} \right)^n + b \left( \frac{1}{2} - \frac{1}{2} \sqrt{5} \right)^n $
As $F_0 = 0$ and $F_1 = 1$, we get
(6a) $ a + b = 0 $
(6b) $ \frac{1}{2} \Big( a + b \Big) + \frac{1}{2} \Big( a - b \Big) \sqrt{5} = 1$
So
(7a) $\displaystyle a = \frac{1}{\sqrt{5}} $
(7b) $\displaystyle a = -\frac{1}{\sqrt{5}} $
Then
(8) $\displaystyle F_n = \frac{ \left( \frac{1}{2} + \frac{1}{2} \sqrt{5} \right)^n - \left( \frac{1}{2} - \frac{1}{2} \sqrt{5} \right)^n }{\sqrt{5}} $
