## Algebra and Trigonometry 10th Edition

Published by Cengage Learning

# Chapter 11 - Review Exercises - Page 840: 2

See below.

#### Work Step by Step

Plugging in the respective values of $n$: $a_1=(-1)^1\frac{5\cdot1}{2\cdot(1-1)+1}=-\frac{5}{1}=-5$ $a_2=(-1)^2\frac{5\cdot2}{2\cdot(2-1)+1}=\frac{10}{3}$ $a_3=(-1)^3\frac{5\cdot3}{2\cdot(3-1)+1}=-\frac{15}{5}=-3$ $a_4=(-1)^4\frac{5\cdot4}{2\cdot(4-1)+1}=\frac{20}{7}$ $a_5=(-1)^5\frac{5\cdot5}{2\cdot(5-1)+1}=-\frac{25}{9}$

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