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

1 Answer
Oct 17, 2017

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

Explanation:

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)