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: 81

Answer

$220$

Work Step by Step

$\sum ^{5}_{i=1}\left( 4i^{2}-2i+6\right) =\sum ^{5}_{i=1}4i^{2}-\sum ^{5}_{i=1}2i+\sum ^{5}_{i=1}6=4\times \dfrac {n\left( n+1\right) \left( 2n+1\right) }{6}-2\times \dfrac {n\left( n+1\right) }{2}+6\times n=4\times \dfrac {5\times \left( 5+1\right) \left( 2\times 5+1\right) }{6}-\dfrac {2\times 5\times \left( 5+1\right) }{2}+6\times 5=220 $
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