Question

find $a_3$.

$a_1 = 1$
$a_n = 8a_{n - 1}$

write your answer as an integer or fraction in simplest form.

$a_3 = \square$

Explanation:

Step1: Find \( a_2 \)

Given \( a_1 = 1 \) and the recurrence relation \( a_n = 8a_{n - 1} \), for \( n = 2 \), we substitute \( n = 2 \) into the recurrence relation. So \( a_2 = 8a_{2 - 1}=8a_1 \). Since \( a_1 = 1 \), then \( a_2 = 8\times1 = 8 \).

Step2: Find \( a_3 \)

Now, for \( n = 3 \), we use the recurrence relation \( a_n = 8a_{n - 1} \) again. Substitute \( n = 3 \), we get \( a_3 = 8a_{3 - 1}=8a_2 \). We already found that \( a_2 = 8 \), so \( a_3 = 8\times8 = 64 \).

Answer:

\( 64 \)