Answer
$18$
Work Step by Step
$\sum ^{3}_{i=1}\left( i+4\right) =\sum ^{3}_{i=1}i+\sum ^{3}_{i=1}4=\dfrac {n\left( n+1\right) }{2}+4n=\dfrac {3\times \left( 3+1\right) }{2}+4\times 3=6+12=18$
Precalculus (6th Edition)
by Lial, Margaret L.; Hornsby, John; Schneider, David I.; Daniels, Callie
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
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