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The given problem involves calculating the sum of an infinite geometric series. The series provided is:

\[
S = \sum_{k=2}^{\infty} 9 \cdot \left( -\frac{1}{5} \right)^k
\]

### Step-by-step Solution:

1. **Recognize the series structure**: The general form of a geometric series is:

   \[
   S = a \sum_{k=0}^{\infty} r^k = \frac{a}{1 - r} \quad \text{for} \quad |r| < 1
   \]

   In this case, the series starts from \( k = 2 \) rather than \( k = 0 \). So we need to adjust for that. The common ratio \( r = -\frac{1}{5} \), and the coefficient outside the summation is \( 9 \).

2. **Factor out the constants**: 
   Factor the constant \( 9 \) out of the summation:

   \[
   S = 9 \sum_{k=2}^{\infty} \left( -\frac{1}{5} \right)^k
   \]

3. **Start the sum from \( k = 0 \)**: 
   The geometric series formula starts at \( k = 0 \), but our series starts at \( k = 2 \). Therefore, we need to subtract the first two terms (i.e., terms for \( k = 0 \) and \( k = 1 \)) from the full sum starting at \( k = 0 \).

   The full sum of the geometric series from \( k = 0 \) is:

   \[
   \sum_{k=0}^{\infty} \left( -\frac{1}{5} \right)^k = \frac{1}{1 - \left( -\frac{1}{5} \right)} = \frac{1}{1 + \frac{1}{5}} = \frac{1}{\frac{6}{5}} = \frac{5}{6}
   \]

4. **Subtract the first two terms**:
   - The \( k=0 \) term is \( \left( -\frac{1}{5} \right)^0 = 1 \),
   - The \( k=1 \) term is \( \left( -\frac{1}{5} \right)^1 = -\frac{1}{5} \).

   So, subtracting these from the full sum:

   \[
   \sum_{k=2}^{\infty} \left( -\frac{1}{5} \right)^k = \frac{5}{6} - 1 + \frac{1}{5}
   \]

   Simplifying this:

   \[
   \frac{5}{6} - 1 = \frac{5}{6} - \frac{6}{6} = -\frac{1}{6}
   \]
   Then:

   \[
   -\frac{1}{6} + \frac{1}{5} = \frac{-5 + 6}{30} = \frac{1}{30}
   \]

5. **Multiply by the factor 9**:

   Finally, multiplying by the coefficient \( 9 \):

   \[
   S = 9 \times \frac{1}{30} = \frac{9}{30} = \frac{3}{10}
   \]

Thus, the sum of the series is **\( \frac{3}{10} \)**, which corresponds to the first option.

### Answer:
\[
\boxed{\frac{3}{10}}
\]

---

Do you want more details on the steps or any clarifications?

Here are five related questions you might find helpful:
1. What is the general formula for a geometric series?
2. How do you handle summations that start at \( k = 1 \) or \( k = 2 \)?
3. What are the conditions for a geometric series to converge?
4. How do you subtract specific terms from an infinite series?
5. How can you apply geometric series in real-world applications?

**Tip**: Always check the convergence of a series before trying to sum it. For a geometric series, it converges when the absolute value of the ratio \( r \) is less than 1.

Given the convergent series \(\sum_{k=2}^{\infty} 9 \cdot \left(-\frac{1}{5}\right)^k\), find the sum of the series.

Learn how to calculate the sum of the convergent geometric series \(\sum_{k=2}^{\infty} 9 \cdot \left(-\frac{1}{5}\right)^k\). This problem covers key concepts in series convergence and geometric series sums, suitable for students in Grades 10-12.

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