Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 10 - Test - Page 1126: 3

Answer

$550$

Work Step by Step

In order to set up the summation notation, we must calculate that value which keeps constantly changing. That value will be designated as the $i$ term. The remaining terms would remain constant in the summation notation. Thus, we get the summation notation as: $\sum_{i=1}^{20} (3i-4)=3 \sum_{i=1}^{20} i+(-4) \sum_{i=1}^{10} 1$ $=3 \sum_{i=1}^{20} i-4(20)$ $=3 \times \dfrac{20 \cdot 21}{2}+80$ $=550$
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