Chapter 8, Sequences and Series - Section 8.1 - Sequences and Summation Notation - 8.1 Exercises - Page 601: 34

$a_{n}=3*(\frac{1}{10})^{n-1}$ Or: $a_{n}=3*(10)^{1-n}$

We are given: $3, 0.3, 0.03, 0.003$ ... We see that each term is the previous term divided by $10$, starting with $3$. We find the pattern: $a_1=3*10^{0}$ $a_2=3*10^{-1}$ $a_3=3*10^{-2}$ $a_4=3*10^{-3}$ Therefore: $a_{n}=3*(\frac{1}{10})^{n-1}$ Or: $a_{n}=3*(10)^{1-n}$