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. 
 A: A proof can be found here involving matrices and eigenvectors.
A: There are many answers here, but none of them address the question of exploting the fact that $\lim_{n \rightarrow \infty}\frac{a_{n+1}}{a_n}=\varphi$. 
To that end, let's begin with a general solution for the Fibonacci sequence with arbitrary initial conditions, $a_0$ and $a_1$. It has already been shown in one of the other answers, so we don't have to derive it. Here it is,
$$a_{n}  = a_{1} \, F_{n} + a_{0} \, F_{n-1}$$
where
$$F_n=\frac{\varphi^n-\psi^n}{\varphi-\psi},\quad \varphi,\psi=\frac{1\pm \sqrt{5}}{2}$$
Now, for sufficiently large $n$, the $\varphi$-term in the numerator dominates, therefore,
$$F_n\approx\frac{\varphi}{\sqrt{5}}$$
We have demonstrated that the following equation will give the solution for $a_n$ for sufficiently large $n$, say $n\gtrapprox6$.
$$a_n=a_1\left\lfloor \frac{\varphi^n}{\sqrt{5}} \right\rceil+a_0\left\lfloor \frac{\varphi^{n-1}}{\sqrt{5}} \right\rceil$$
where the brackets indicate the nearest integer (i.e., rounding). This has been verified numerically for random $a_0$ and $a_1$.
A: Let $ F_n$ denote the $n$th term of the fibonacci series,
and $\phi=\dfrac{1+\sqrt{5}}{2}$.
Then, $$F_n=\dfrac{\phi^n -(1-\phi)^n}{\sqrt5}$$.
For its proof you can refer to this site http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibformproof.html .
A: 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. 
A: You know:
$\begin{align*}
  F_n
   &= \frac{1}{\sqrt{5}} \left( \phi^n - (1 - \phi)^n \right)
\end{align*}$
where $\phi = (1 + \sqrt{5}) / 2$. As:
$\begin{align*}
  \frac{1 - \phi}{2 \sqrt{5}}
    &= -0.1382
\end{align*}$
so the second term is much less than $1/2$ for all $n \ge 1$, you can just take the above formula and round:
$\begin{align*}
F_n
  &= \left\lfloor \frac{\phi^n}{\sqrt{5}} + \frac{1}{2} \right\rfloor
\end{align*}$
Fun, but next to useless (to compute $\phi$ to the required number of digits to get $F_n$ for a large $n$ is more work than using other methods).
A: To begin with, generate a vector $u_k=\begin{bmatrix}F_{k+1}\\F_k\end{bmatrix}$.
It is not hard to obtain the Fibonacci matrix, from $F_{k+2}=F_{k+1}+F_k$, denoted as $A=\begin{bmatrix}1&1\\1&0\end{bmatrix}$, such that $u_{k+1}=Au_{k}$.
By mathematical induction, it can be proved that $u_k=A^ku_0$, where $u_0=\begin{bmatrix}1\\0\end{bmatrix}$. $u_k=A^ku_0$ is easy to calculate after diagolized as $A^k=\begin{bmatrix}\lambda_1&\lambda_2\\1&1\end{bmatrix}\begin{bmatrix}\lambda_1&\\&\lambda_2\end{bmatrix}^k\begin{bmatrix}\lambda_1&\lambda_2\\1&1\end{bmatrix}^{-1}$, which uses eigenvalues of $A$ that are$\lambda_1=\frac{1+\sqrt5}{2},\lambda_2=\frac{1-\sqrt5}{2}$, and the eigenvectors. Pick the second entry of $u_k$, $F_k$ can be obtained:
$$ F_k=\frac{1}{\lambda_1-\lambda_2}(\lambda_1^k-\lambda_2^k)=\frac{1}{\sqrt 5}[(\frac{1+\sqrt5}{2})^k-(\frac{1-\sqrt5}{2})^k]
$$
