Question

已知等比数列的首项a1=1,公比为q,前n项和为Sn，设q≥2，bn=S2/S1+S3/S2+...+Sn+1/Sn，bn≥3n。求q的最小值

Answer 1

an=q^(n-1)
Sn=(q^n-1)/(q-1)
S(n+1)/Sn=1+a(n+1)/Sn
=1+(q-1)q^n/(q^n-1)
=1+(q-1)/[1-(1/q)^n]

又q≥2
0＜1/q≤1/2＜1
0＜(1/q)^(n+1)≤(1/q)^n≤(1/q)^(n-1)≤1/2
0＞-(1/q)^(n+1)≥-(1/q)^n≥-(1/q)^(n-1)≥-1/2
1＞1-(1/q)^(n+1)≥1-(1/q)^n≥1-(1/q)^(n-1)≥1/2＞0
1＜1/[1-(1/q)^(n+1)]≤1/[1-(1/q)^n]≤1/[1-(1/q)^(n-1)]≤2
S(n+1)/Sn=1+(q-1)/[1-(1/q)^n]
≥1+(q-1)/[1-(1/q)^(n+1)]

bn={1+(q-1)/[1-(1/q)^n]}+{1+(q-1)/[1-(1/q)^(n-1)]}+{1+(q-1)/[1-(1/q)^(n-2)]}+{1+(q-1)/[1-(1/q)^(n-3)]}+……+{1+(q-1)/[1-(1/q)^3]}+{1+(q-1)/[1-(1/q)^2]}+{1+(q-1)/[1-(1/q)]}
=n+(q-1)/[1-(1/q)^n]+(q-1)/[1-(1/q)^(n-1)]+(q-1)/[1-(1/q)^(n-2)]+(q-1)/[1-(1/q)^(n-3)]+……+(q-1)/[1-(1/q)^3]+(q-1)/[1-(1/q)^2]+(q-1)/[1-(1/q)]
≥n+n(q-1)/[1-(1/q)^n]
≥3n
(q-1)/[1-(1/q)^n]≥2
q-1≥2-2(1/q)^n
q^(n+1)-3q^n+2≥0

q^2-3q+2≥0,q≥2