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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.

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  • $\begingroup$ The characterisitic equation is $\lambda^2 - \lambda - 1 = 0$ so $2\lambda_{1,2} = -1 \pm \sqrt{5}$. Then our solution is $\alpha \lambda_1 + \beta \lambda_2$. We can find $\alpha$ and $\beta$ in terms of $a_0$ and $a_1$ by solving a $2\times 2$ system. The fibonnaci sequence can then be found by using the suitable values of $a_{0,1}$. $\endgroup$ – Zain Patel May 23 '17 at 18:34
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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 $a_1$.

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A proof can be found here involving matrices and eigenvectors.

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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 .

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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.

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