## College Algebra 7th Edition

Published by Brooks Cole

# Chapter 8, Sequences and Series - Section 8.1 - Sequences and Summation Notation - 8.1 Exercises - Page 601: 75

#### Answer

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

#### Work Step by Step

We are given: $\sqrt{2}, \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}, \sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}}$, ... We write each terms as a power of $2$: $a_1=\sqrt{2}=2^{1/2}$ $a_{2}=\sqrt{2^1*2^{1/2}}=\sqrt{2^{3/2}}=2^{3/4}$ $a_{3}=\sqrt{2^1*2^{3/4}}=\sqrt{2^{7/4}}=2^{7/8}$ $a_{4}=\sqrt{2^1*2^{7/8}}=\sqrt{2^{15/8}}=2^{15/16}$ We see that the denominator in the power doubles (powers of 2) and the numerator is the denominator minus 1. Therefore: $a_{n}=2^{(2^{n}-1)/2^{n}}$

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