This problem involves finding the sum of a geometric series. Let's analyze the given series:

$S = -2 + (-6) + (-18) + \cdots + (-162)$

Step 1: Verify the terms

• First term ($a$) = $-2$
• Common ratio ($r$) = $\frac{-6}{-2} = 3$

The general formula for the sum of a geometric series is:

$S_n = a \frac{1 - r^n}{1 - r}, \quad r \neq 1$

Step 2: Find the number of terms ($n$)

The last term ($a_n$) is given as $-162$. The formula for the $n$-th term in a geometric series is:

$a_n = a \cdot r^{n-1}$

Substituting $a = -2$, $r = 3$, and $a_n = -162$:

$-162 = -2 \cdot 3^{n-1}$

Divide both sides by $-2$:

$81 = 3^{n-1}$

Solve for $n-1$:

$n-1 = \log_3{81} = 4$

So, $n = 5$.

Step 3: Calculate the sum

Using the formula for the sum of the first $n$ terms:

$S_n = -2 \frac{1 - 3^5}{1 - 3}$

Simplify the denominator:

$1 - 3 = -2$

Simplify the numerator:

$3^5 = 243, \quad 1 - 243 = -242$

Substitute back into the formula:

$S_5 = -2 \frac{-242}{-2}$

$S_5 = -242$

Final Answer:

The partial sum is $-242$.

Would you like a detailed explanation for any specific part?

Here are 5 related questions to explore further:

1. What is the formula for the sum of an infinite geometric series?
2. How do you determine whether a geometric series converges or diverges?
3. How would the result change if the common ratio were a fraction?
4. Can a geometric series have alternating signs? How would it affect the sum?
5. How do you derive the formula for the $n$-th term of a geometric sequence?

Tip: Always double-check the number of terms when calculating the sum of a finite series.