The value of $$3 + \cfrac{1}{4 + \cfrac{1}{3 + \cfrac{1}{4 + \cfrac{1}{3 + \ldots \infty}}}}$$ is equal to:

Let the value of the infinite continued fraction be $$x$$. Since the pattern alternates between 3 and 4, we can write: $$x = 3 + \cfrac{1}{4 + \cfrac{1}{x}}$$, because after the first $$3 + \frac{1}{4 + \cdots}$$, the pattern repeats from 3 again.

Simplifying the inner fraction: $$x = 3 + \frac{1}{4 + \frac{1}{x}} = 3 + \frac{1}{\frac{4x + 1}{x}} = 3 + \frac{x}{4x + 1}$$.

So $$x = \frac{3(4x + 1) + x}{4x + 1} = \frac{12x + 3 + x}{4x + 1} = \frac{13x + 3}{4x + 1}$$.

Cross-multiplying: $$x(4x + 1) = 13x + 3$$, which gives $$4x^2 + x = 13x + 3$$, or $$4x^2 - 12x - 3 = 0$$.

Using the quadratic formula: $$x = \frac{12 \pm \sqrt{144 + 48}}{8} = \frac{12 \pm \sqrt{192}}{8} = \frac{12 \pm 8\sqrt{3}}{8} = \frac{3 \pm 2\sqrt{3}}{2}$$.

Since $$x$$ must be positive and $$\frac{3 - 2\sqrt{3}}{2} = \frac{3 - 3.464}{2} < 0$$, we take the positive root: $$x = \frac{3 + 2\sqrt{3}}{2} = 1.5 + \sqrt{3}$$.

This matches Option A: $$1.5 + \sqrt{3}$$.