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: 84

Answer

$973$

Work Step by Step

$\sum ^{6}_{i=1}\left( i^{2}+2i^{3}\right) =\sum ^{6}_{i=1}i^{2}+\sum ^{6}_{i=1}2i^{3}=\dfrac {n\left( n+1\right) \left( 2n+1\right) }{6}+\dfrac {2\times n^{2}\left( n+1\right) ^{2}}{4}=\dfrac {6\times \left( 6+1\right) \left( 2\times 6+1\right) }{6}+\dfrac {2\times 6^{2}\times \left( 6+1\right) ^{2}}{4}=973 $
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