Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 11 - Section 11.1 - Sequences and Summation Notation - Exercise Set - Page 829: 21



Work Step by Step

Since, we have $\sum_{k=1}^{5}k(k+4)=1(5)+2(6)+3(7)+4(8)+5(9)$ or, $=5+12+21+32+45$ Hence, we get $=115$
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