Answer
$3+4+...+(n+2)$
Work Step by Step
$\sum_{k=1}^{n} (k+2)=(1+2)+(2+2)+...+(n+2)=3+4+...+(n+2)$
Precalculus (10th Edition)
by Sullivan, Michael
Published by
Pearson
ISBN 10:
0-32197-907-9
ISBN 13:
978-0-32197-907-0
Chapter 12 - Sequences; Induction; the Binomial Theorem - 12.1 Sequences - 12.1 Assess Your Understanding - Page 807: 49
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