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The problem is evaluating the sum of a geometric series:

\[
\sum_{n=1}^6 3 \left(\frac{3}{2}\right)^{n-1}
\]

For a geometric series, the formula is given as:

\[
S = a \frac{1 - r^n}{1 - r}
\]

Where:
- \( a \) is the first term,
- \( r \) is the common ratio, and
- \( n \) is the number of terms.

### Step-by-Step Solution:
1. **Identify \( a \), \( r \), and \( n \):**
   - \( a = 3 \)
   - \( r = \frac{3}{2} \)
   - \( n = 6 \)

2. **Substitute these values into the formula:**
   \[
   S = 3 \cdot \frac{1 - \left(\frac{3}{2}\right)^6}{1 - \frac{3}{2}}
   \]

3. **Simplify the denominator:**
   \[
   1 - \frac{3}{2} = -\frac{1}{2}
   \]

   So:
   \[
   S = 3 \cdot \frac{1 - \left(\frac{3}{2}\right)^6}{-\frac{1}{2}}
   \]

4. **Simplify the division:**
   Dividing by \(-\frac{1}{2}\) is the same as multiplying by \(-2\), so:
   \[
   S = -6 \cdot \left(1 - \left(\frac{3}{2}\right)^6\right)
   \]

5. **Calculate \(\left(\frac{3}{2}\right)^6\):**
   Using the value:
   \[
   \left(\frac{3}{2}\right)^6 = \left(1.5\right)^6 = 11.390625
   \]

6. **Substitute back:**
   \[
   S = -6 \cdot \left(1 - 11.390625\right)
   \]

7. **Simplify further:**
   \[
   S = -6 \cdot (-10.390625)
   \]
   \[
   S = 62.34375
   \]

8. **Round to the nearest hundredth:**
   \[
   S \approx 62.34
   \]

### Final Answer:
\[
S = 62.34
\]

Would you like me to elaborate on any step or verify further? Let me know!

---

### Related Questions:
1. How do you determine the first term and common ratio of a geometric sequence?
2. What is the general formula for finding the sum of an infinite geometric series?
3. How does rounding affect the accuracy of your final answer in mathematical computations?
4. What happens if the common ratio \( r \) is greater than 1 in a geometric series?
5. Can you provide examples where geometric series appear in real-life applications?

### Tip:
Always double-check your work by substituting the calculated sum back into the original equation to verify its accuracy!

Evaluate the geometric series: S = ∑ (from n = 1 to 6) [3 * (3/2)^(n-1)]. Round to the nearest hundredth.

Learn how to evaluate the geometric series S = ∑ 3 * (3/2)^(n-1) for n = 1 to 6 using the geometric series formula. Step-by-step instructions include identifying the first term, ratio, and number of terms, as well as rounding to the nearest hundredth. Perfect for Grades 9-12.

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