Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 10 - Section 10.1 - Sequences and Summation Notation - Exercise Set - Page 1048: 39



Work Step by Step

Here, we have $\sum_{i=0}^4\dfrac{(-1)^i}{i!}=\dfrac{(-1)^0}{0!}+\dfrac{(-1)^1}{1!}+\dfrac{(-1)^2}{2!}+\dfrac{(-1)^3}{3!}+\dfrac{(-1)^4}{4!}$ $=\dfrac{1}{1}+(\dfrac{-1}{1})+\dfrac{1}{2}+(\dfrac{-1}{6})+(\dfrac{1}{24})$ $=\dfrac{12}{24}-\dfrac{4}{24}+\dfrac{1}{24}$ $\dfrac{9}{24}$ $=\dfrac{3}{8}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.