Chapter 9 - Section 9.1 - Sequences - Exercises - Page 487: 1

$a_1=0$ $a_2=\frac{-1}{4}$ $a_3=\frac{-2}{9}$ $a_4=\frac{-3}{16}$

We know that $a_n=\frac{1-n}{n^2}$. Hence, here: $a_1=\frac{1-1}{1^2}=\frac{0}{1}=0$ $a_2=\frac{1-2}{2^2}=\frac{-1}{4}$ $a_3=\frac{1-3}{3^2}=\frac{-2}{9}$ $a_4=\frac{1-4}{4^2}=\frac{-3}{16}$