Golden Sum

Because Bob recently posted an article linking golden ratio to Fibonacci numbers

This motivates me to find $4180$ and $6764$ in terms of Fibonacci numbers. Turns out $F_{19} = 4181, F_{20} = 6765$

With the identity: $\phi^n = F_{n-1} + F_n \phi$ , and $\phi^2 - \phi - 1 = 0$

$\begin{aligned} F_{19} + F_{20} \phi & = & \phi^{20} \\ (F_{19} - 1) + (F_{20} - 1)\phi & = & \phi^{20} - 1 - \phi \\ 4180 + 6764 \phi & = & \phi^{20} - (1 + \phi) \\ \displaystyle \sum_{i=1}^N \phi^i & = & \phi^{20} - \phi^2 \\ \frac {\phi \left ( \phi^N - 1 \right ) }{\phi - 1} & = & \phi^{20} - \phi^2 \\ \frac {\phi^N - 1 }{\phi - 1} & = & \phi^{19} - \phi \\ \phi^N - 1 & = & (\phi^{19} - \phi)(\phi - 1 ) \\ & = & \phi (\phi^{18} - 1)(\phi - 1 ) \\ & = & (\phi^{18} - 1)(\phi^2 - \phi ) \\ & = & (\phi^{18} - 1)(1) \\ N & = & \boxed{18} \\ \end{aligned}$

Fistly,

$\phi^1=\phi$

And for $N\geq2$ ,

$\phi^N=F_N\phi+F_{N-1}$

where $F_N$ is the N-th Fibonacci number.

Given that

$\displaystyle \sum_{i=1}^N \phi^i=6764\phi+4180$

Then clearly we have

$F_1+F_2+...+F_{N-1}+F_N=6764$

$F_1+F_2+...+F_{N-1}=4180$

Subtract these two equation we have

$F_N=2584$

Then $N=18$ .