Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.1 - Sequences - Exercises 10.1 - Page 570: 99

Answer

$x_n=2^{n-2}$; for all $n \geq 2$

Work Step by Step

According to the Principle of Mathematical Induction $a_n \geq 0$. This implies that there exist $n$ for all natural numbers N. Given: $x_{n+1}=x_1+x_2+...+x_n$ and $x_1=1$ and, $x_2=x_1=1$; $x_3=x_1+x_2=1+1=2$ and $x_4=x_1+x_2+x_3=1+1+2=4....$ In the same manner, we have $x_1=1,x_2=2^0,x_3=2^1....$ We can see that the pattern we follow for sequence $x_n$ will be: $x_n=2^{n-2}$; for all $n \geq 2$
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