Solving using geometric Series, #sqrt(2)/2, 1/2, 2^(3/2)/8,1/4#?

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Answer 1 · 2017-10-17T02:10:50

#a_n = sqrt2/2(sqrt2/2)^(n-1)#

A geometric sequence is

#a, ar, ar^2, ar^3,cdots #

Given the sequence:

#sqrt(2)/2, 1/2, 2^(3/2)/8,1/4#

#a = sqrt2/2#

#ar = 1/2#

#(ar)/a = r#

#r = (1/2)/(sqrt2/2)#

#r = 1/sqrt2#

#r = sqrt2/2#

#a_n = sqrt2/2(sqrt2/2)^(n-1)#