To find the sum of the series $\sum_{k=1}^\infty \frac{3n - 2}{5^n}$, we will approach it by splitting it into two separate sums and then finding each of those sums individually.

Let's rewrite the given series as: $\sum_{n=1}^\infty \frac{3n - 2}{5^n} = \sum_{n=1}^\infty \frac{3n}{5^n} - \sum_{n=1}^\infty \frac{2}{5^n}$

First, we evaluate the second sum: $\sum_{n=1}^\infty \frac{2}{5^n}$ This is a geometric series with the first term $a = \frac{2}{5}$ and common ratio $r = \frac{1}{5}$. The sum of an infinite geometric series is given by: $S = \frac{a}{1 - r} = \frac{\frac{2}{5}}{1 - \frac{1}{5}} = \frac{\frac{2}{5}}{\frac{4}{5}} = \frac{2}{4} = \frac{1}{2}$

Now, we need to evaluate the first sum: $\sum_{n=1}^\infty \frac{3n}{5^n}$ We use the formula for the sum of a series involving $n$ multiplied by a geometric progression: $\sum_{n=1}^\infty \frac{n}{5^n} = \frac{r}{(1 - r)^2}$ where $r = \frac{1}{5}$. Therefore: $\sum_{n=1}^\infty \frac{n}{5^n} = \frac{\frac{1}{5}}{(1 - \frac{1}{5})^2} = \frac{\frac{1}{5}}{(\frac{4}{5})^2} = \frac{\frac{1}{5}}{\frac{16}{25}} = \frac{25}{20} = \frac{5}{4}$

Therefore, we have: $\sum_{n=1}^\infty \frac{3n}{5^n} = 3 \times \frac{5}{4} = \frac{15}{4}$

Putting it all together: $\sum_{n=1}^\infty \frac{3n - 2}{5^n} = \frac{15}{4} - \frac{1}{2} = \frac{15}{4} - \frac{2}{4} = \frac{13}{4}$

Thus, the sum of the series is: $\boxed{\frac{13}{4}}$

Do you want more details or have any questions?

Here are 5 more questions you can ask:

1. How do you find the sum of a geometric series?
2. Can you explain the steps to derive the sum formula for $\sum_{n=1}^\infty \frac{n}{5^n}$?
3. What are other methods to solve series involving factorials?
4. How do you approach solving an alternating series?
5. Can you explain the concept of convergence in series?

Tip: When dealing with series, breaking them down into simpler parts can often simplify the problem-solving process.