## Algebra 2 (1st Edition)

$a_n=a_1 r^{n-1}=8^{n-1}$ and $S_{8}=2396745$
The nth term is given by $a_n= a_1 r^{n-1}$ ...(1) Here, we have: $a_1=1$, $a_2=8$ and $a_n=a_1 r^{n-1}=8^{n-1}$ We know that $S_{n}=a_1(\dfrac{1-r^{n}}{1-r})$ Now, $S_{8}=(1) \times (\dfrac{1-8^{8}}{1-8})$ This gives: $S_{8}=2396745$ Hence, $a_n=a_1 r^{n-1}=8^{n-1}$ and $S_{8}=2396745$