Question

等比数列{an}的前n项和为Sn,已知S1.S3.S2.成等差数列,(2)若a1-a3=-3/2,求{n.an}的前n项和

Answer 1

解：
设公比为q
S1、S3、S2成等差数列，则
2S3=S1+S2
2(a1+a1q+a1q²)=a1+a1+a1q
整理，得
2q²+q=0
q(2q+1)=0
q=0(舍去)或q=-1/2
a1-a3=-3/2
a1-a1q²=-3/2
(1-q²)a1=-3/2
a1=(-3/2)/(1-q²)=(-3/2)/[1-(-1/2)²]=-2
an=a1q^(n-1)=(-2)×(-1/2)^(n-1)=(-1/2)^(n-2)
nan=n×(-1/2)^(n-2)
Tn=a1+2a2+...+nan=1×(-1/2)^(-1) +2×(-1/2)^0+3×(-1/2)+...+n×(-1/2)^(n-2)
(-1/2)Tn=1×(-1/2)^0+2×(-1/2)+...+(n-1)×(-1/2)^(n-2)+n×(-1/2)^(n-1)
Tn-(-1/2)Tn=(3/2)Tn=(-1/2)^(-1)+(-1/2)^0+...+(-1/2)^(n-2) -n×(-1/2)^(n-1)
=(-2)×[1-(-1/2)ⁿ]/[1-(-1/2)] -n×(-1/2)^(n-1)
=[(4+6n)/3](-1/2)ⁿ -4/3
Tn=[(8+12n)×(-1/2)ⁿ -8]/9