Let's solve each problem step by step.

Part I: Find the sum of each finite geometric sequence.

The formula to find the sum $S_n$ of a finite geometric sequence is:

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

• $a$ is the first term,
• $r$ is the common ratio, and
• $n$ is the number of terms.

1. $1, 2, 4, 8, \dots, n = 6$

• First term $a = 1$,
• Common ratio $r = \frac{2}{1} = 2$,
• Number of terms $n = 6$.

Using the formula:

$S_6 = \frac{1(1 - 2^6)}{1 - 2} = \frac{1(1 - 64)}{-1} = 63$

2. $2, -10, 50, -250, \dots, n = 8$

• First term $a = 2$,
• Common ratio $r = \frac{-10}{2} = -5$,
• Number of terms $n = 8$.

Using the formula:

$S_8 = \frac{2(1 - (-5)^8)}{1 - (-5)} = \frac{2(1 - 390625)}{6} = \frac{2(-390624)}{6} = -130208$

3. $1, -4, 16, -64, \dots, n = 9$

• First term $a = 1$,
• Common ratio $r = \frac{-4}{1} = -4$,
• Number of terms $n = 9$.

Using the formula:

$S_9 = \frac{1(1 - (-4)^9)}{1 - (-4)} = \frac{1(1 + 262144)}{5} = \frac{262145}{5} = 52429$

4. $-2, -6, -18, -54, \dots, n = 9$

• First term $a = -2$,
• Common ratio $r = \frac{-6}{-2} = 3$,
• Number of terms $n = 9$.

Using the formula:

$S_9 = \frac{-2(1 - 3^9)}{1 - 3} = \frac{-2(1 - 19683)}{-2} = \frac{-2(-19682)}{-2} = 19682$

5. $1, -5, 25, -125, \dots, n = 7$

• First term $a = 1$,
• Common ratio $r = \frac{-5}{1} = -5$,
• Number of terms $n = 7$.

Using the formula:

$S_7 = \frac{1(1 - (-5)^7)}{1 - (-5)} = \frac{1(1 + 78125)}{6} = \frac{78126}{6} = 13021$

Part II: Find the sum of each infinite geometric sequence.

The sum $S$ of an infinite geometric sequence can be found using the formula:

$S = \frac{a}{1 - r}$ where $|r| < 1$.

1. $1, 0.5, 0.25, 0.125, \dots$

• First term $a = 1$,
• Common ratio $r = \frac{0.5}{1} = 0.5$.

Using the formula:

$S = \frac{1}{1 - 0.5} = \frac{1}{0.5} = 2$

2. $3, \frac{9}{4}, \frac{27}{16}, \frac{81}{64}, \dots$

• First term $a = 3$,
• Common ratio $r = \frac{\frac{9}{4}}{3} = \frac{3}{4}$.

Using the formula:

$S = \frac{3}{1 - \frac{3}{4}} = \frac{3}{\frac{1}{4}} = 12$

3. $81, -27, 9, -3, \dots$

• First term $a = 81$,
• Common ratio $r = \frac{-27}{81} = -\frac{1}{3}$.

Using the formula:

$S = \frac{81}{1 - \left(-\frac{1}{3}\right)} = \frac{81}{\frac{4}{3}} = \frac{81 \times 3}{4} = 60.75$

4. $1, -0.6, 0.36, -0.216, \dots$

• First term $a = 1$,
• Common ratio $r = \frac{-0.6}{1} = -0.6$.

Using the formula:

$S = \frac{1}{1 - (-0.6)} = \frac{1}{1.6} = 0.625$

5. $5, \frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \dots$

• First term $a = 5$,
• Common ratio $r = \frac{\frac{5}{2}}{5} = \frac{1}{2}$.

Using the formula:

$S = \frac{5}{1 - \frac{1}{2}} = \frac{5}{\frac{1}{2}} = 10$

Do you have any questions or would you like further details on any of the steps?

Here are five related questions to practice:

1. How do you determine the common ratio in a geometric sequence?
2. What is the condition for an infinite geometric series to have a sum?
3. How does the number of terms $n$ affect the sum of a finite geometric sequence?
4. Can you derive the sum formula for a finite geometric sequence from the partial sums method?
5. How do you find the sum of a geometric sequence when the first term is negative?

Tip: Always check if the common ratio's absolute value is less than 1 when solving for the sum of an infinite geometric sequence.