Precalculus (6th Edition) Blitzer

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

Chapter 10 - Section 10.1 - Sequences and Summation Notation - Exercise Set - Page 1048: 36



Work Step by Step

Here, we have $\sum_{i=2}^4(\dfrac{-1}{3})=(\dfrac{-1}{3})^2+(\dfrac{-1}{3})^3+(\dfrac{-1}{3})^4$ $=\dfrac{1}{9}+(\dfrac{-1}{27})+\dfrac{1}{81}$ $=\dfrac{9-3+1}{81}$ $=\dfrac{7}{81}$
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