Precalculus (6th Edition)

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

Chapter 11 - Further Topics in Algebra - 11.2 Arithmetic Sequences and Series - 11.2 Exercises - Page 1026: 67



Work Step by Step

$\sum ^{3}_{i=1}\left( i+4\right) =\sum ^{3}_{i=1}i+\sum ^{3}_{i=1}4=\dfrac {n\left( n+1\right) }{2}+4n=\dfrac {3\times \left( 3+1\right) }{2}+4\times 3=6+12=18$
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.