Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 12 - Section 12.1 - Sequences and Summation Notation - 12.1 Exercises - Page 850: 12

Answer

$a_{1}=\displaystyle \frac{1}{2}$, $a_{2}=-\displaystyle \frac{2}{3}$, $a_{3}=\displaystyle \frac{3}{4}$, $a_{4}=-\displaystyle \frac{4}{5}$, $a_{100}=-\displaystyle \frac{100}{101}$

Work Step by Step

$a_{n}=\displaystyle \frac{(-1)^{n+1}n}{n+1}$. $a_{1}=\displaystyle \frac{(-1)^{2}\cdot 1}{1+1}=\frac{1}{2}$, $a_{2}=\displaystyle \frac{(-1)^{3}\cdot 2}{2+1}=-\frac{2}{3}$, $a_{3}=\displaystyle \frac{(-1)^{4}\cdot 3}{3+1}=\frac{3}{4}$, $a_{4}=\displaystyle \frac{(-1)^{5}\cdot 4}{4+1}=-\frac{4}{5}$, $a_{100}=\displaystyle \frac{(-1)^{101}\cdot 100}{101}=-\frac{100}{101}$
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