If we label the pds across successive vertical resistors "$V_n$", "$V_{n+1}$" and so on we can write the Kirchoff current law at a junction as$$\frac{V_n-V_{n+1}}{R_A}=\frac{V_{n+1}}{R_B}+\frac{V_{n+1}-V_{n+2}}{R_A}$$
in which $R_A$ and $R_B$ are the values of a horizontal and vertical resistor respectively. If you're happy with the assumption that $\frac{V_{n+2}}{V_{n+1}}=\frac{V_{n+1}}{V_{n}}\ [=\alpha\ \,\text{say}]\ $ throughout the ladder, then you have a quadratic equation for $\alpha$. Knowing $\alpha$ you can find the input current for a given input pd, and hence the input resistance.
Possibly rather less convincing than the 'no change with an extra section' method, but at least it's rather different! And it does give the same answer!
A few numerical investigations… Suppose the right hand end of the ladder is a vertical resistor (2 ohm) and the last section of the ladder is a horizontal resistor (1 ohm) connected to that vertical resistor. Then the resistance, $R_1$, of the last section seen from the left is 3 ohm. The resistance, $R_2$, of the last two sections together is, seen from the left, 11/5 ohm, then we have 43/21 ohm, 171/85 ohm. These resistances, I found, fit the pattern$$R_{m}\text{\ohm}=2+\frac{3}{4^{m}-1}.$$Clearly the convergence to 2 ohm will be very rapid!
Addition to post
Simplify notation by calling the horizontal resistance "X" and the vertical resistance "$\beta X$". Call the resistance of a ladder of m sections (one horizontal resistor joined on the right to a vertical resistor down to a 'ground' rail), "$r_{m}X$". Then, assuming that adding one more section on to the left of an infinite ladder will make no difference to its resistance we have$$r_{\infty}X=X+\frac{\beta X.r_{\infty}X}{\beta X+r_{\infty}X}\ \ \ \text{giving}\ \ \ \ \ r_{\infty}=\frac{1}{2}\left(1+\sqrt{1+4\beta}\right)$$
$\beta=2$ gives $r_{\infty}=2$, as we know. Certain other values of $\beta$ will also give rational values of $r_{\infty}$, for example, $\beta \ =$12, 6, 2, 3/4, 5/16, 9/64 give respectively, $r_{\infty}=\ $4, 3, 2, 3/2, 5/4, 9/8. No doubt in each of these cases we can find a general formula for $r_m$ as we did for when $\beta=2$, by inspecting the bit left over when we subtract $r_\infty$ from $r_1$, $r_2$ and so on.
For most values of $\beta$, this procedure won't stand a chance of working, because $r_{\infty}$ will be irrational, while $r_1$, $r_2$ and so on are clearly rational if $\beta$ is rational! Instead, for the general case, we use binomial series... Expanding the expression for $r_\infty$ given above we get
$$r_\infty=1+\beta-\beta^2+2\beta^3-5\beta^4+14\beta^5-42\beta^6+132\beta^7-429\beta^8+...$$
But clearly, $r_1=1+\beta$.
So $r_1$ agrees with $r_\infty$ up to the second term!
We find without much difficulty that$$r_2=1+\frac{\beta(1+\beta)}{1+2\beta}$$
Expanding$\frac{1}{1+2\beta}$ binomially and multiplying out, we get$$r_2=1+\beta-\beta^2 +2\beta^3 -4\beta^4+...$$
So we now have agreement with $r_\infty$ up to the fourth term!
Continuing…$$r_3=1+\frac{\beta(1+3\beta +\beta^2)}{1+4\beta +3\beta^2}$$and this gives us$$r_3=1+\beta-\beta^2+2\beta^3-5\beta^4+14\beta^5-41\beta^6+…$$
So we now have agreement with $r_\infty$ up to the sixth term!
This may not have been very elegant, but I'm certainly now convinced! [Presumably $\beta$ is restricted to being less than 1 for this method to be valid.] Thank you for this interesting problem.
If two resistors are connected in parallel and one of them is 0 that is a short circuit. All the current will flow through the short circuit $R_1$ (the 0-ohm resistor) and none of it through the other one $R_2$.
However, stting a resistor to $0$ will give you some issues and indeterminate forms like $0/0$ and $0\infty$. So you need to choose some conditions to solve them. The easiest thing to imagine is that, if there is no resistence there is no drop in potential hence $V=V_1=V_2=0$.
For the same reason, as the resistor $r$ is shorted, there is no drop in voltage across it and hence, the $V$ in your equation is $V=0$. In symbols,
$V_2=0$ and hence $I_2=V_2/R_2=0$ or $V_2=R_2I_2=0$. No problem here because $R_2$ is finite and we just "assumed" $V_2=0$ because there is no drop in potential as the parallel has $0$ resistence in total.
Instead, for resistor $R_1$ you get $V_1=V_2=0$ (because the two resistors are in parallel)
$I_1=V_1/R_1=0/0$ and $V_1=R_1I_1=0\infty$ which are indeterminate forms - but you know $V_1$ is 0 because you have shorted the circuit by definition: there is no resistance so there is no drop in voltage.
Notice that you get indeterminate forms... that also holds for the total current $I_t=V/R_t=0/\infty$. In real circuits (except superconductors) the resistence is never 0, even a shorted wire has some (very low) resistence, so there always is a finite a current and a finite (small) drop in voltage so that $R_1=0$ is impossible, and one always has $R_1=r_0$ with $r_0$ possibly small but non-zero.
Best Answer
The resistance $R = \frac{U}{I}$.
The inversed resistance $G = \frac{1}{R} = \frac{I}{U}$ is the electrical conductance .
Resistances are additive in serial scenarios, as voltages at the same current are additive.
$$R_\mathrm{T} = \frac{U_\mathrm{T}}{I} = \frac {U_1 + U_2 + U_3}{I} = \frac{U_1}{I} + \frac{U_2}{I} + \frac{U_3}{I} = R_1 + R_2 + R_3$$
Conductances are additive in parallel scenarios, as currents at the same voltage are additive.
$$G_\mathrm{T} = \frac{I_\mathrm{T}}{U} = \frac{I_1 + I_2 + I_3}{U} = \frac{I_1}{U} + \frac{I_2}{U} + \frac{I_3}{U} = G_1 + G_2 + G_3$$
If there are parallel resistors 1 Ohm, 2 Ohm, 5 Ohm, they have the respective conductances 1 S, 0.5 S, 0.2 S, with the summary conductance 1.7 S, which is equivalent to the resistance 1/1.7 Ohm.
For $n$ general resistors:
$$R_\mathrm{T} = \frac 1{G_\mathrm{T}} = \frac {1}{ \sum_{i=1}^{n} G_i } = \frac {1}{ \sum_{i=1}^{n} \frac{1}{R_i} }$$
It is valid even for other passive components like capacitors and inductors, if the respective complex arithmetic and generalized quantities impedance (complex resistance) and admittance (complex conductance) are involved.
Few analogies: