Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 12 - Sequences; Induction; the Binomial Theorem - Chapter Review - Review Exercises - Page 838: 11


Geometric Sum: $6\left(1-\dfrac{1}{2^n}\right)$

Work Step by Step

We are given the sequence: $3,\dfrac{3}{2},\dfrac{3}{4},\dfrac{3}{8},\dfrac{3}{16},.....$ We determine the ratio between consecutive terms: $\dfrac{a_2}{a_1}=\dfrac{\dfrac{3}{2}}{3}=\dfrac{1}{2}$ $\dfrac{a_3}{a_2}=\dfrac{\dfrac{3}{4}}{\dfrac{3}{2}}=\dfrac{1}{2}$ $\dfrac{a_4}{a_3}=\dfrac{\dfrac{3}{8}}{\dfrac{3}{4}}=\dfrac{1}{2}$ $\dfrac{a_5}{a_4}=\dfrac{\dfrac{3}{16}}{\dfrac{3}{8}}=\dfrac{1}{2}$ As the ratio between consecutive terms is constant, the sequence is geometric. Its elements are: $a_1=3$ $r=\dfrac{1}{2}$ Determine the sum of the first $n$ terms: $S_n=a_1\cdot\dfrac{1-r^n}{1-r}=3\cdot\dfrac{1-\left(\dfrac{1}{2}\right)^n}{1-\dfrac{1}{2}}=3\left(1-\dfrac{1}{2^n}\right)\cdot 2=6\left(1-\dfrac{1}{2^n}\right)$
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