The sum to the infinity of the series $$1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + \ldots$$ is

Write the given infinite series as

$$S = 1 + \frac{2}{3} + \frac{6}{3^{2}} + \frac{10}{3^{3}} + \frac{14}{3^{4}} + \ldots$$

From the numerators we observe
$$1,\,2,\,6,\,10,\,14,\ldots$$
After the first two terms the numbers increase by $$4$$ each time. Hence for $$n \ge 2$$ the numerator can be written as $$4n-2$$, where $$n$$ is the power of $$3$$ in the denominator.

Therefore split the series:

$$S = 1 + \frac{2}{3} + \sum_{n=2}^{\infty} \frac{4n-2}{3^{n}}$$

Separate the two sums inside the sigma:

$$\sum_{n=2}^{\infty} \frac{4n-2}{3^{n}} = 4\sum_{n=2}^{\infty} \frac{n}{3^{n}} \;-\; 2\sum_{n=2}^{\infty} \frac{1}{3^{n}}$$

Let $$x = \frac13$$. We need the standard power-series results (valid for $$|x|\lt1$$):
$$\sum_{n=0}^{\infty} x^{n} = \frac{1}{1-x},$$
$$\sum_{n=0}^{\infty} n x^{n} = \frac{x}{(1-x)^{2}}.$$

1. Geometric tail beginning with $$n=2$$: $$\sum_{n=2}^{\infty} x^{n} = \frac{x^{2}}{1-x}.$$ Putting $$x=\tfrac13$$ gives $$\sum_{n=2}^{\infty} \frac{1}{3^{n}} = \frac{(1/3)^{2}}{1-\tfrac13} = \frac{1/9}{2/3} = \frac16.$$

2. Arithmetico-geometric tail beginning with $$n=2$$: Start with the complete sum and subtract the $$n=1$$ term (the $$n=0$$ term is zero):
$$\sum_{n=2}^{\infty} n x^{n} = \frac{x}{(1-x)^{2}} - x.$$ Putting $$x=\tfrac13$$ gives $$\sum_{n=2}^{\infty} \frac{n}{3^{n}} = \frac{\tfrac13}{(2/3)^{2}} - \frac13 = \frac{1/3}{4/9} - \frac13 = \frac{3}{4} - \frac13 = \frac{5}{12}.$$

Now assemble the two parts:

$$\sum_{n=2}^{\infty} \frac{4n-2}{3^{n}} = 4\left(\frac{5}{12}\right) - 2\left(\frac16\right) = \frac{20}{12} - \frac{2}{6} = \frac{5}{3} - \frac13 = \frac{4}{3}.$$

Add the first two terms of the original series:

$$S = 1 + \frac{2}{3} + \frac{4}{3} = 1 + 2 = 3.$$

Therefore, the sum to infinity of the series is $$3$$.

Option B which is: $$3$$