College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 8 - Sequences, Induction, and Probability - Exercise Set 8.1 - Page 714: 40

Answer

$\frac{19}{30}$

Work Step by Step

given i = 0 to 4 in $\frac{(-1)^{i + 1}}{(i + 1)!}$ Then the sum = $\frac{(-1)^{0 + 1}}{(0 + 1)!}$ + $\frac{(-1)^{1 + 1}}{(1 + 1)!}$ +$\frac{(-1)^{2 + 1}}{(2 + 1)!}$ + $\frac{(-1)^{3 + 1}}{(3 + 1)!}$ + $\frac{(-1)^{4 + 1}}{(4 + 1)!}$ = $\frac{(-1)^{1}}{(1)!}$ + $\frac{(-1)^{2}}{(2)!}$ +$\frac{(-1)^{3}}{(3)!}$ + $\frac{(-1)^{4}}{(4)!}$ + $\frac{(-1)^{5}}{(5)!}$ = -1 + $\frac{1}{2}$ +$\frac{(-1)}{6}$ + $\frac{1}{24}$ + $\frac{(-1)}{120}$ = -$\frac{1}{2}$ -$\frac{1}{6}$ + $\frac{1}{24}$ - $\frac{1}{120}$ = $\frac{- 60 - 20 + 5 - 1}{120}$ = $\frac{- 76}{120}$ = $-\frac{76}{120}$ = $\frac{19}{30}$
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