Question

已知正项等差数列{an}的前n项和为Sn,若S3=12,且2a1,a2,a3+1成等比数列 （1）记bn=an/3^n的前n项和为Tn,求T

求Tn
答案：5/4-(6n+5)/(4×3^n)

Answer 1

解：
设数列公差为d，数列为正项数列，则d≥0
S3=3a1+3d=12
a2=a1+d=4
a2^2=2a1(a3+1)
a2^2=2(a2-d)(a2+d+1)
2(4-d)(4+d+1)=16
整理，得
d^2+d-12=0
(d+4)(d-3)=0
d=3或d=-4(<0，舍去)
a1=a2-d=4-3=1
an=1+(n-1)×3=3n-2
bn=an/3^n=(3n-2)/3^n=n/3^(n-1)-2/3^n
令kn=n/3^(n-1)，则其n项和Kn=1/1+2/3+3/3^2+4/3^3+...+(n-1)/3^(n-2)+n/3^(n-1) (1)
Kn/3=1/3+2/3^2+3/3^3+...+(n-1)/3^(n-1)+n/3^n (2)
(1)-(2)
(2/3)Kn=1+2/3-1/3+1/3^2+1/3^3+...+1/3^(n-1)-n/3^n
=1+[1/3+1/3^2+1/3^3+...+1/3^(n-1)]-n/3^n
=1+(1/3)[1-(1/3)^n]/(1-1/3)-n/3^n
=1+(1/2)-(1/2)(1/3)^n-n/3^n
=3/2-(2n+1)/(2×3^n)
Kn=9/4-(6n+3)/(4×3^n)
-2/3^n=(-2/3)[1-(1/3)^n]/(1-1/3)=1/3^n-1
Tn=Kn+1/3^n-1
=9/4-(6n+3)/(4×3^n)+1/3^n-1
=5/4-(6n+3-4)/(4×3^n)
=5/4-(6n-1)/(4×3^n)