Algebra 2 (1st Edition)

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

Chapter 12 Sequences and Series - Chapter Test - Page 843: 15



Work Step by Step

We know that $\sum_{i=1}^n(i)=\frac{n(n+1)}{2}$ Hence $\sum_{i=1}^{10}(4i-9)=10(-9)+4\sum_{i=1}^{10}(i)=-90+4\frac{10(10+1)}{2}=130$
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