Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 12 Sequences and Series - Cumulative Review - Page 848: 38



Work Step by Step

We know that $\sum_{i=1}^n(i)=\frac{n(n+1)}{2}$ Hence $\sum_{i=1}^{16}(-2+i)=16(-2)+\sum_{i=1}^{16}(i)=-32+\frac{16(16+1)}{2}=104$
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