Golden Ratio

We'll generalize it , think of the quadratic equation whose roots are

$\color{#3D99F6}{\alpha=\dfrac{1+\sqrt{5}}{2}}$ and $\color{#69047E}{\beta = \dfrac{1-\sqrt{5}}{2}}$

The quadratic equation will be $\color{#D61F06}{x^2-x-1=0}$

Now, make recurrence relation from this quadratic equation, such that $\color{#20A900}{a_n= \alpha ^n +\beta ^n}$

This is possible, see that if you make the quadratic to become the characteristic equation, then recurrence will be $\color{#20A900}{a_n-a_{n-1}-a_{n-2}=0}$ , or more simple looking, $\color{#20A900}{a_n=a_{n-1}+a_{n-2}}$

And, now , by subsituting $n=0$ and $n=1$ , for general form, we get

$a_0= \alpha^0 + \beta ^0= 1+1=2$

$a_1= \alpha+\beta = 1$

Now, from the recurrence, we conclude that add adjacent two terms to get the next get third term, add 2nd and 3rd to get 4th, and so on, so the sequence is $2,1,3,4,7,11,18,29,47,76,123$ , giving $10^{th}$ term as $\boxed{\color{#D61F06}{123}}$

This sequence is actually Luca's sequence.

Note that $\varphi = \dfrac {1+\sqrt 5}2$ is the golden ratio, $\dfrac {1-\sqrt 5}2 = - \dfrac 1\varphi$ , and $\varphi + \left(- \dfrac 1\varphi \right) = 1$ . Then we have

$\begin{aligned} \left(\varphi - \frac 1\varphi \right)^2 & = 1^2 \\ \varphi^2 - 2 + \frac 1{\varphi^2} & = 1 \\ \implies \varphi^2 + \frac 1{\varphi^2} & = 3 \\ \left(\varphi^2 + \frac 1{\varphi^2}\right)\left(\varphi - \frac 1\varphi \right) & = 3 \times 1 \\ \varphi^3 - \varphi + \frac 1\varphi - \frac 1{\varphi^2} & = 3 \\ \implies \varphi^3 - \frac 1{\varphi^3} & = 4 \\ \left(\varphi^3 - \frac 1{\varphi^3}\right)\left(\varphi^2 + \frac 1{\varphi^2} \right) & = 4 \times 3 \\ \varphi^5 + \varphi - \frac 1\varphi - \frac 1{\varphi^5} & = 12 \\ \implies \varphi^5 - \frac 1{\varphi^5} & = 11 \\ \left(\varphi^5 - \frac 1{\varphi^5}\right)^2 & = 11^2 \\ \varphi^{10} -2 + \frac 1{\varphi^{10}} & = 121 \\ \implies \varphi^{10} +\left(- \frac 1\varphi \right)^{10} & = \boxed{123} \end{aligned}$