Answer
Geometric
Sum: $\dfrac{8}{7}(8^n-1)$
Work Step by Step
We are given the sequence:
$s_n=\{2^{3n}\}$
We determine the ratio between consecutive terms:
$\dfrac{s_{n+1}}{s_n}=\dfrac{2^{3(n+1)}}{2^{3n}}=\dfrac{2^{3n}\cdot 2^3}{2^{3n}}=8$
As the ratio between consecutive terms is constant, the sequence is geometric. Its elements are:
$s_1=2^{3(1)}=8$
$r=8$
Determine the sum of the first $n$ terms:
$S_n=a_1\cdot\dfrac{1-r^n}{1-r}=8\cdot\dfrac{1-8^n}{1-8}=\dfrac{8}{7}(8^n-1)$
