## Precalculus (10th Edition)

$\dfrac{1024}{5}$
An infinite geometric series converges if and only if $|r|\lt1$, where $r$ is the common ratio. If it converges, then its sum equals $\frac{a_1}{1-r}$ where $a_1$ is the first term. The common ratio is the quotient of two consecutive terms. Thus, the given series has: $r=\dfrac{a_2}{a_1}=\dfrac{-64}{256}=-\dfrac{1}{4}$ $\left|-\dfrac{1}{4}\right|=\dfrac{1}{4}\lt1$, thus the series converges. Hence, with $a_1=256$, the sum is: $\dfrac{256}{1-(-\frac{1}{4})}=\dfrac{256}{1+\frac{1}{4}}=\dfrac{256}{\frac{5}{4}}=256 \cdot \dfrac{4}{5}=\dfrac{1024}{5}$