已知｛an｝是等比数列a2＝2,a5＝1╱4,则Sn＝a1+a2+...+an的取值范围?

Question

已知｛an｝是等比数列a2＝2,a5＝1╱4,则Sn＝a1+a2+...+an的取值范围?

Answer 1

an=a1.q^(n-1)
a2=2
a1q=2 (1)
a5 = 1/4
a1.q^4 = 1/4 (2)
(2)/(1)
q^3 = 1/8
q = 1/2
from (1) => a1= 4
an = 4.(1/2)^(n-1)
Sn =a1+a2+...+an
= 8( 1- (1/2)^n )
< 8
Sn ≥ S1 = 4
ie
4≤ Sn