Answer
$115$
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
$=115$
Intermediate Algebra for College Students (7th Edition)
by Blitzer, Robert F.
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
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