Let’s see another way to get the same result as (2), which should provide another reason why the “golden number” is appearing in the equivalent resistance expression.
Let’s build the equivalent resistance step by step. Let’s recall we’re now considering the net made by all equale resistors ($R$)
In the first step, let’s calculate the equivalent resistance as if the net was made of the first 2 resistors only. It is:
$R_{EQ,1}=2R=\frac{2}{1}R$
In the second step, let’s add one couple of resistors (one series and one parallel). We have:
$R_{EQ,2}=R+R||2R=R+\frac{2R^{2}}{3R}=\frac{5}{3}R$
In the third step, let’s add one more couple of resistors (one series and one parallel). We have:
$R_{EQ,3}=R+R||(R+R||2R)=R+R||\frac{5}{3}R=R+\frac{(\frac{5}{3}R^{2})}{\frac{8}{3}R}=R+\frac{5}{8}R=\frac{13}{8}R$
Going on few steps, we’ll get the following sequence:
$\left \{R_{EQ,n}\right \}=\left \{ \frac{2}{1}R, \frac{5}{3}R, \frac{13}{8}R, \frac{34}{21}R, … \right \}$
This should remind us the Fibonacci’s sequence:
$f_{n}=\left \{ 1,1,2,3,5,8,13,21,34,55,… \right \}$
In fact, we have
$\left \{R_{EQ,n}\right \}=\left \{ \frac{f_{2n+1}}{f_{2n}}R \right \}$ (3)
Now, since the sequence (3) is a sub-sequence of $\left \{ \frac{f_{n+1}}{f_{n}}R \right \}$ (i.e. the sequence of ratios of 2 adjacent Fibonacci’s sequence terms) whose limit is the golden number, we have:
$R_{EQ}=\lim_{n}R_{EQ,n}=\lim_{n}\frac{f_{2n+1}}{f_{2n}}R=\lim_{n}\frac{f_{n+1}}{f_{n}}R=\frac{1+\sqrt{5}}{2}R$,
which is the same result as (2).