Calculus: Early Transcendentals 8th Edition

$$a_n = -\frac{(-1)^n}{4^{n-2}}$$
Our first term is $a_1 = 4$. Because the sequence alternates, we have a $(-1)^{n}$ in the sequence. Each term also changes by a factor of $\frac{1}{4}$, so we introduce a factor of $(\frac{1}{4})^n$. Putting together the pieces with the fact that $a_1 = 4$, we have: $$a_n = -\frac{(-1)^n}{4^{n-2}}$$