What is the general form of the sequence What is the general form of the sequence describing an infinite resistor circuit as shown
The sequence can be shown as: 2R,  5R/3,  8R/13,  13R/21,  ...

 A: Looking at the physics of it, if $R_n$ is the resistance at step n, then $R_1$ = 2R as stated.
At step $n+1$ you add one resistor in series and one in parallel, so that $R_{n + 1} = R + 1 / (1/ R + 1/R_n)$. 
This gives $R_2 = R + 1/(1/R + 1/2R) = R(1 + 2/3) = 5R/3$, and 
$R_3 = R + 1/(1/R + 1/(5R/3)) = 13R/8$
$R_4 = R + 1/(1/R + 1/(13R/8)) = 34R/21$
So, that's a recursive formula for the terms of the series. Perhaps someone can resolve it as an expression for $R_n$ as a function of n: I regret that I can't. The expression can be simplified somewhat. Let $R_n = R.a_n$ then the terms for $a_n$ are defined by $a_1$ = 2; $a_{n+1} = 1 + [a_n/(1 + a_n)]$
A: The sequence is generated by the ratio of successive Fibonacci numbers:
$$R_n = R\frac{F_{2n+1}}{F_{2n}}$$
$$F_1 = 1, F_2=1,\\F_{n+1}=F_n+F_{n-1}$$
The Fibonacci numbers have the closed form representation
$$F_n = \frac{\varphi^{n}-(1-\varphi)^{n}}{\sqrt{5}}$$
where $\varphi = (1 + \sqrt{5}) /2 $ 
is the Golden ratio.
Hence,
$$R_n = R\frac{\varphi^{2n+1}-(1-\varphi)^{2n+1}}{\varphi^{2n}-(1-\varphi)^{2n}}$$
Also we would find from circuit analysis, that the nth stage resistance is related to the nth convergent of the continued fraction representation of $\varphi$.
$$\varphi = 1+\frac{1}{1+\frac{1}{1+...}}$$
