# Chapter 12 - Sequences; Induction; the Binomial Theorem - 12.1 Sequences - 12.1 Assess Your Understanding - Page 807: 48

$2^{1/2},2^{-1/4},2^{-5/8},2^{-13/16},2^{-29/32}$

#### Work Step by Step

We are given the sequence: $a_1=\sqrt 2$ $a_n=\sqrt{\dfrac{a_{n-1}}{2}}$ Determine the first 5 terms of the sequence by substituting 1, 2, 3, 4, 5 for $n$: $a_1=\sqrt 2$ $a_2=\sqrt{\dfrac{a_1}{2}}=\sqrt{\dfrac{\sqrt 2}{2}}=\sqrt {2^{-1/2}}=2^{-1/4}$ $a_3=\sqrt{\dfrac{a_2}{2}}=\sqrt{\dfrac{2^{-1/4}}{2}}=\sqrt{2^{-5/4}}=2^{-5/8}$ $a_4=\sqrt{\dfrac{a_3}{2}}=\sqrt{\dfrac{2^{-5/8}}{2}}=\sqrt{2^{-13/8}}=2^{-13/16}$ $a_5=\sqrt{\dfrac{a_4}{2}}=\sqrt{\dfrac{2^{-13/16}}{2}}=\sqrt{2^{-29/16}}=2^{-29/32}$

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