University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.2 - Sigma Notation and Limits of Finite Sums - Exercises: 1

Answer

$7$

Work Step by Step

$\sum\limits_{k=1}^2\frac{6k}{k+1}=\frac{6\cdot1}{1+1}+\frac{6\cdot2}{2+1}=\frac{6}{2}+\frac{12}{3}=3+4=7$
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