Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 11 - Further Topics in Algebra - 11.1 Sequences and Series - 11.1 Exercises - Page 1015: 67

Answer

$124+111+100+91$ $=426$

Work Step by Step

$\sum_{i=1}^4\frac{{x_i}^3+1000}{x_i+10}$ $=\frac{{x_1}^3+1000}{x_1+10}+\frac{{x_2}^3+1000}{x_2+10}+\frac{{x_3}^3+1000}{x_3+10}+\frac{{x_4}^3+1000}{x_4+10}$ $=\frac{(-2)^3+1000}{-2+10}+\frac{(-1)^3+1000}{-1+10}+\frac{0^3+1000}{0+10}+\frac{1^3+1000}{1+10}$ $=\frac{992}{8}+\frac{999}{9}+\frac{1000}{10}+\frac{1001}{11}$ $=124+111+100+91$ $=\boxed{426}$
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.