## Precalculus (10th Edition)

Geometric; $-\dfrac{2}{3}(4^n-1)$
We are given the sequence: $\left\{-\dfrac{1}{2}\cdot 4^n\right\}$ Determine the ratio of consecutive terms: $\dfrac{a_{n+1}}{a_n}=\dfrac{-\dfrac{1}{2}\cdot 4^{n+1}}{-\dfrac{1}{2}\cdot 4^n}=4$ As the ratio of consecutive terms is constant, the sequence is geometric. Its elements are: $a_1=-\dfrac{1}{2}\cdot 4^1=-\dfrac{1}{2}\cdot 4=-2$ $r=4$ Determine the sum of the first $n$ terms: $S_n=a_1\dfrac{1-r^n}{1-r}=-2\cdot\dfrac{1-4^n}{1-4}=-\dfrac{2}{3}(4^n-1)$