Question

已知{an}是公比为q的等比数列，且a1，a3，a2成等差数列．

Answer 1

1.

a1+a1*q=2a1*q^2

2q^2-q-1=0

q=-1/2或q=1

2.

(1)q=-1/2

bn=2-(n-1)/2

Sn=(9n-n^2)/4>=2

1<=n<=8

Sn-bn=(-n^2+11n-10)/4>=0

∴Sn>=bn

(2)q=1

bn=2+(n-1)

Sn=(n^2+3n)/2>=2

则n>=1

Sn-bn=(n^2+n-2)>=0

∴Sn>=bn