The sum of the first 5 terms of a geometric progression is given by S5 = 2(1−(135)). Identify the typo and calculate the correct sum.

The "13" in the expression $S_5 = 2(1-(13^5))$ is a typo.

Based on the context of a geometric progression and the subsequent correct calculation steps, the common ratio $r$ is $\frac{1}{3}$. Therefore, the term $(13^5)$ should actually be $(\frac{1}{3})^5$.

Let's correct the line and show the intended calculation:

The formula for the sum of the first $n$ terms of a geometric progression is:

$$S_n = \frac{a(1-r^n)}{1-r}$$

Given $a=2$, $r=\frac{1}{3}$, and $n=5$:

$$S_5 = \frac{2(1-(\frac{1}{3})^5)}{1-\frac{1}{3}}$$ $$S_5 = \frac{2(1-\frac{1^5}{3^5})}{\frac{3}{3}-\frac{1}{3}}$$ $$S_5 = \frac{2(1-\frac{1}{243})}{\frac{2}{3}}$$ $$S_5 = \frac{2(\frac{243-1}{243})}{\frac{2}{3}}$$ $$S_5 = \frac{2(\frac{242}{243})}{\frac{2}{3}}$$

To divide by a fraction, multiply by its reciprocal:

$$S_5 = 2 \times \frac{242}{243} \times \frac{3}{2}$$ $$S_5 = \frac{2 \times 242 \times 3}{243 \times 2}$$ $$S_5 = \frac{242 \times 3}{243}$$ $$S_5 = \frac{726}{243}$$ $$S_5 = \frac{242}{81}$$

The final answer is $\frac{242}{81}$.