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 - 12.2 Arithmetic Sequences - 12.2 Assess Your Understanding - Page 814: 16

Answer

We know that $e^{\ln{n}}=n$, In order for a sequence to be arithmetic, the difference of all consecutive terms must be constant. Hence here: $a_{n+1}-a_n=(e^{\ln{n+1}})-e^{\ln{n}}=(n+1)-(n)=1$, thus it is an arithmetic sequence. $a_1=1$ $a_2=2$ $a_3=3$ $a_4=4$

Work Step by Step

We know that $e^{\ln{n}}=n$, In order for a sequence to be arithmetic, the difference of all consecutive terms must be constant. Hence here: $a_{n+1}-a_n=(e^{\ln{n+1}})-e^{\ln{n}}=(n+1)-(n)=1$, thus it is an arithmetic sequence. $a_1=1$ $a_2=2$ $a_3=3$ $a_4=4$
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