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: 83

Answer

$304$

Work Step by Step

$\sum ^{4}_{i=1}\left( 3i^{3}+2i-4\right) =\sum ^{4}_{i=1}3i^{3}+\sum ^{4}_{i=1}2i-\sum ^{4}_{i=1}4=3\times \dfrac {n^{2}\left( n+1\right) ^{2}}{4}+2\times \dfrac {n\left( n+1\right) }{2}-4\times n=\dfrac {3\times 4^{2}\times \left( 4+1\right) ^{2}}{4}+2\times \dfrac {4\times \left( 4+1\right) }{2}-4\times 4=304$
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