Consider an infinite ladder constructed using 1\ and 2\ resistors,...

Let the effective resistance of the infinite ladder between points A and B be $R$.

Since the ladder is infinite, removing the first repeating unit (one $1\ \Omega$ resistor in series and one $2\ \Omega$ resistor in parallel) leaves an identical infinite ladder across the next nodes. Therefore, the resistance of the remaining infinite ladder is also $R$.

The equivalent circuit can be simplified to a $1\ \Omega$ resistor in series with the parallel combination of a $2\ \Omega$ resistor and $R$.

The equation for the equivalent resistance is:
$R = 1 + \dfrac{2R}{2 + R}$

Solving for $R$:
$R = \dfrac{2 + R + 2R}{2 + R}$

$R = \dfrac{3R + 2}{R + 2}$

$R(R + 2) = 3R + 2$

$R^2 + 2R = 3R + 2$

$R^2 - R - 2 = 0$

Factorizing the quadratic equation:
$(R - 2)(R + 1) = 0$

Since resistance cannot be negative, we discard $R = -1\ \Omega$.

Thus, $R = 2\ \Omega$.