Question

evaluate each geometric series described.

1. $1 + 5 + 25 + 125..., n = 6$

1. $-4 - 16 - 64 - 256..., n = 9$

1. $2 + 6 + 18 + 54..., n = 8$

1. $-3 - 12 - 48 - 192..., n = 8$

1. $4 + 8 + 16 + 32..., n = 8$

Explanation:

Problem 11: \( 1 + 5 + 25 + 125..., n = 6 \)

Step 1: Identify \( a_1 \), \( r \), and \( n \)

For a geometric series, the first term \( a_1 = 1 \). The common ratio \( r \) is found by dividing a term by its previous term: \( \frac{5}{1} = 5 \). The number of terms \( n = 6 \).

Step 2: Use the geometric series sum formula

The formula for the sum of the first \( n \) terms of a geometric series is \( S_n = \frac{a_1(r^n - 1)}{r - 1} \) (when \( r
eq 1 \)). Substituting \( a_1 = 1 \), \( r = 5 \), and \( n = 6 \):

$$ S_6 = \frac{1(5^6 - 1)}{5 - 1} $$

Step 3: Calculate \( 5^6 \) and simplify

\( 5^6 = 15625 \). So,

$$ S_6 = \frac{15625 - 1}{4} = \frac{15624}{4} = 3906 $$

Step 1: Identify \( a_1 \), \( r \), and \( n \)

First term \( a_1 = -4 \). Common ratio \( r = \frac{-16}{-4} = 4 \). Number of terms \( n = 9 \).

Step 2: Use the geometric series sum formula

Using \( S_n = \frac{a_1(r^n - 1)}{r - 1} \) (since \( r
eq 1 \)):

$$ S_9 = \frac{-4(4^9 - 1)}{4 - 1} $$

Step 3: Calculate \( 4^9 \) and simplify

\( 4^9 = 262144 \). Then,

$$ S_9 = \frac{-4(262144 - 1)}{3} = \frac{-4(262143)}{3} = -4 \times 87381 = -349524 $$

Step 1: Identify \( a_1 \), \( r \), and \( n \)

First term \( a_1 = 2 \). Common ratio \( r = \frac{6}{2} = 3 \). Number of terms \( n = 8 \).

Step 2: Use the geometric series sum formula

Using \( S_n = \frac{a_1(r^n - 1)}{r - 1} \):

$$ S_8 = \frac{2(3^8 - 1)}{3 - 1} $$

Step 3: Calculate \( 3^8 \) and simplify

\( 3^8 = 6561 \). So,

$$ S_8 = \frac{2(6561 - 1)}{2} = 6561 - 1 = 6560 $$

Answer:

\( 3906 \)

Problem 13: \( -4 - 16 - 64 - 256..., n = 9 \)