College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 9 - Section 9.1 - Sequences - 9.1 Assess Your Understanding - Page 647: 52

Answer

$4+9+16+25+36+\cdots+(n+1)^{2}$

Work Step by Step

There are n terms, as the index k changes from 1 to n. The index k dictates how the terms are formed: $\displaystyle \sum_{k=1}^{n}(k+1)^{2}=(1+1)^{2}+(2+1)^{2}+(3+1)^{2}+...+(n+1)^{2}$ $=4+9+16+25+36+\cdots+(n+1)^{2}$
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