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 - Quiz for Lessons 12.1-12.3 - Page 817: 13


$a_n=2 (4^{n-1})$ $a_{15}=536870912$ $S_{15}=715827882$

Work Step by Step

The general formula for the nth term of a geometric series is given by $a_n= a_1r^{n-1}$ ...(1) The ratio of the successive terms is $4$. Equation (1) gives: $a_n= a_1r^{n-1}=2 \times 4^{n-1}$ Plugging in $n =15$, we have $a_{15}=2 \times 4^{15-1}=536870912$ We know that $S_{n}=\dfrac{a_1(1-r^n)}{1-r}$ Now, $S_{15}=\dfrac{2(1-4^{15})}{1-4}=715827882$
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