The sum of the infinite series $$1 + \frac{5}{6} + \frac{12}{6^2} + \frac{22}{6^3} + \frac{35}{6^4} + \frac{51}{6^5} + \frac{70}{6^6} + \ldots$$ is equal to:

We need to find the sum: $$S = 1 + \frac{5}{6} + \frac{12}{6^2} + \frac{22}{6^3} + \frac{35}{6^4} + \frac{51}{6^5} + \frac{70}{6^6} + \ldots$$ The numerators are 1, 5, 12, 22, 35, 51, 70, ... whose first differences 4, 7, 10, 13, 16, 19, ... form an arithmetic progression with common difference 3, and whose constant second differences 3 imply a quadratic general term: $$a_n = An^2 + Bn + C$$.

Using $$a_1 = 1, a_2 = 5, a_3 = 12$$ leads to the system $$A + B + C = 1,\quad 4A + 2B + C = 5,\quad 9A + 3B + C = 12$$, which simplifies to $$3A + B = 4,\quad 5A + B = 7$$ and yields $$A = \frac{3}{2},\; B = -\frac{1}{2},\; C = 0$$. Thus $$a_n = \frac{3n^2 - n}{2} = \frac{n(3n-1)}{2}\;.$$

Writing the series with $$r = \frac{1}{6}$$ gives $$S = \sum_{n=1}^{\infty} \frac{n(3n-1)}{2} \left(\frac{1}{6}\right)^{n-1} = \frac{3}{2}\sum_{n=1}^{\infty}n^2 r^{n-1} - \frac{1}{2}\sum_{n=1}^{\infty}n\,r^{n-1}$$ where $$r = \frac{1}{6}$$.

Using the standard formulas $$\sum_{n=1}^{\infty} n r^{n-1} = \frac{1}{(1-r)^2} = \frac{1}{(5/6)^2} = \frac{36}{25}$$ and $$\sum_{n=1}^{\infty} n^2 r^{n-1} = \frac{1+r}{(1-r)^3} = \frac{1+1/6}{(5/6)^3} = \frac{7/6}{125/216} = \frac{7}{6} \times \frac{216}{125} = \frac{252}{125}\;,$$ we obtain $$S = \frac{3}{2}\times\frac{252}{125} - \frac{1}{2}\times\frac{36}{25} = \frac{756}{250} - \frac{36}{50} = \frac{756}{250} - \frac{180}{250} = \frac{576}{250} = \frac{288}{125}\;.$$

Therefore, the answer is Option C: $$\boldsymbol{\frac{288}{125}}$$.