We can decompose the entire network into three parallel components connected across terminals $A$ and $B$:

1. The central vertical resistor $R$ connecting $A$ and $B$.
2. An infinite section extending to the right ($+\hat{i}$).
3. An infinite section extending to the left ($-\hat{i}$).

Let $R_1$ be the equivalent resistance of one semi-infinite section (e.g., the right side) as seen from the junction points. Since the ladder is infinite, adding one more “unit cell” to the front of the ladder does not change its equivalent resistance.

The unit cell consists of two series resistors (top and bottom rails, each $R$) and one shunt resistor (vertical rung, $R$). Therefore, we can write the equation for $R_1$:

$$ R_1 = R_{\text{series\_top}} + R_{\text{series\_bot}} + (R_{\text{shunt}} \parallel R_1) $$ $$ R_1 = R + R + \left( \frac{R \cdot R_1}{R + R_1} \right) $$ $$ R_1 = 2R + \frac{RR_1}{R + R_1} $$

Multiplying both sides by $(R + R_1)$ to clear the denominator:

$$ R_1(R + R_1) = 2R(R + R_1) + RR_1 $$ $$ RR_1 + R_1^2 = 2R^2 + 2RR_1 + RR_1 $$ $$ R_1^2 + RR_1 = 2R^2 + 3RR_1 $$

Rearranging into a standard quadratic form $ax^2 + bx + c = 0$:

$$ R_1^2 – 2RR_1 – 2R^2 = 0 $$

Using the quadratic formula to solve for $R_1$:

$$ R_1 = \frac{-(-2R) \pm \sqrt{(-2R)^2 – 4(1)(-2R^2)}}{2} $$ $$ R_1 = \frac{2R \pm \sqrt{4R^2 + 8R^2}}{2} $$ $$ R_1 = \frac{2R \pm \sqrt{12R^2}}{2} = \frac{2R \pm 2R\sqrt{3}}{2} $$

Since resistance must be positive, we take the positive root:

$$ R_1 = R(1 + \sqrt{3}) $$

The total resistance $R_{eq}$ measured between terminals $A$ and $B$ is the parallel combination of three branches:

1. The Left Semi-infinite section ($R_1$)
2. The Right Semi-infinite section ($R_1$)
3. The Central Vertical resistor ($R$)