Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 12 - Section 12.1 - Sequences and Summation Notation - 12.1 Exercises: 75

Answer

$a_{n}=2^{(2^{n}-1)/2^{n}}$

Work Step by Step

$k=1, a_{1}=2^{1/2}$ $k=2, a_{2}=(2\cdot 2^{1/2})^{1/2}=(2^{3/2})^{1/2}=2^{3/4}$ $k=3, a_{3}=(2a_{2}^{1/2})^{1/2}=(2\cdot 2^{3/4})^{1/2}=(2^{7/4})^{1/2}=2^{7/8}$ $k=4, a_{3}=(2a_{3}^{1/2})^{1/2}=(2\cdot 2^{7/8})^{1/2}=(2^{15/4})^{1/2}=2^{15/16}$ The pattern that is emerging for the exponent is k=1, $ \ \ 1=2^{1}-1, \ \ 2=2^{1}$ k=2, $ \ \ 3=2^{2}-1, \ \ 4=2^{2}$ k=3, $ \ \ 7=2^{3}-1, \ \ 8=2^{3}$ k=4, $ \ \ 15=2^{4}-1, \ \ 16=2^{4}$ So, for $a_{n}, $ n-th numerator=$2^{n}-1$ n-th denominator$= 2^{n}$ $a_{n}=2^{(2^{n}-1)/2^{n}}$
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