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: 67

Answer

$18$

Work Step by Step

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