题目

已知数列an的前n项和为Sn,且向量a等于（n,Sn）,向量b等于（1,n）公线,（1）求证数列an是等差数列（2）若bn=3的n次方分之an,求数列bn的前n项和Tn

提问时间：2020-12-15

答案

(1)
Sn/n = n/1
Sn = n^2
a1=1
an = Sn-S(n-1)
= 2n-1
=>{an}是等差数列
(2)
bn = an/3^n
= 2(n.(1/3)^n) - 1/3^n
Tn=b1+b2+...+bn
=2S - (1/2)(1- 1/3^n)
S = 1.(1/3)^1+2(1/3)^2+...+n.(1/3)^n (1)
(1/2)S = 1.(1/3)^2+2(1/3)^3+...+n.(1/3)^(n+1) (2)
(1)-(2)
(1/2)S =(1/3+1/3^2+...+1/3^n) -n.(1/3)^(n+1)
= (1/2)( 1- 1/3^n) - n.(1/3)^(n+1)
S = 2 - (2n+3).(1/3)^(n+1)
Tn=2S - (1/2)(1- 1/3^n)
=4 - 2(2n+3).(1/3)^(n+1) - (1/2)(1- 1/3^n)
=(1/2)[ 7 - (8n+15)(1/3)^(n+1) ]