Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 11 - Sequences; Induction; the Binomial Theorem - Section 11.3 Geometric Sequences; Geometric Series - 11.3 Assess Your Understanding - Page 844: 43


$S_n=2[1-( \dfrac{2}{3})^n]$

Work Step by Step

The sum of the First $n$ Terms of a Geometric Sequence is given by: $S_{n}=\displaystyle\sum_{k=1}^n a_1r^{k-1}=a_{1} (\dfrac{1-r^{n}}{1-r}) ; \ r\neq 0,1$ We are given: $a_{1}=\dfrac{2}{3} ; \ r= \dfrac{2}{3}$ Thus, we can write the sum as: $\displaystyle\sum_{k=1}^n (\dfrac{2}{3})^{k}=\displaystyle\sum_{k=1}^n (\dfrac{2}{3}) (\dfrac{2}{3})^{k-1}$ Now, $S_n= \dfrac{2}{3} \ [\dfrac{1-(\dfrac{2}{3} )^{n}}{1- \dfrac{2}{3} } \ ] \\=\dfrac{2}{3} \ [\dfrac{1-(\dfrac{2}{3} )^{n}}{ \dfrac{1}{3} } \ ] $ Therefore, $S_n=2[1-( \dfrac{2}{3})^n]$
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