Algebra 2 (1st Edition)

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

Chapter 12 Sequences and Series - 12.3 Analyze Geometric Sequences and Series - 12.3 Exercises - Skill Practice - Page 815: 49



Work Step by Step

Here, we have $\sum_{i=1}^8 6(4)^{i-1}=6\sum_{i=1}^8 (4)^{i-1}$ We know that $S_{n}=a_1(\dfrac{1-r^{n}}{1-r})$ Now, $\sum_{i=1}^8 6(4)^{i-1}=6\sum_{i=1}^8 (4)^{i-1}=1(\dfrac{1-4^{8}}{1-4})$ or, $=6 \times (\dfrac{1-65536}{-3})$ or, $=131070$
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