The sum of the infinite series $$1 + \frac{2}{3} + \frac{7}{3^2} + \frac{12}{3^3} + \frac{17}{3^4} + \frac{22}{3^5} + \ldots$$ is equal to:

The given series is $$S = 1 + \frac{2}{3} + \frac{7}{3^2} + \frac{12}{3^3} + \frac{17}{3^4} + \frac{22}{3^5} + \ldots$$

The numerators after the first term are 2, 7, 12, 17, 22, ... which form an arithmetic progression with first term 2 and common difference 5. The general term for $$n \geq 1$$ is $$5n - 3$$. So we can write $$S = 1 + \sum_{n=1}^{\infty} \frac{5n - 3}{3^n}$$.

We use the standard results: $$\sum_{n=1}^{\infty} \frac{n}{3^n} = \frac{3}{(3-1)^2} = \frac{3}{4}$$ and $$\sum_{n=1}^{\infty} \frac{1}{3^n} = \frac{1}{2}$$.

Therefore $$S = 1 + 5 \cdot \frac{3}{4} - 3 \cdot \frac{1}{2} = 1 + \frac{15}{4} - \frac{3}{2} = 1 + \frac{15}{4} - \frac{6}{4} = 1 + \frac{9}{4} = \frac{13}{4}$$.