## Precalculus (10th Edition)

Geometric Sum: $\dfrac{8}{7}(8^n-1)$
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: $S_n=a_1\cdot\dfrac{1-r^n}{1-r}=8\cdot\dfrac{1-8^n}{1-8}=\dfrac{8}{7}(8^n-1)$