Question

设数列{an}的前n项和为Sn,点(an,Sn)在直线y=3/2x-1上

求数列{an}的通项公式
在an与a（n+1）之间插入n个数，使这n+2个数组成公差为dn的等差数列，求数列{1/dn}的前n项和Tn

Answer 1

(an,Sn)在直线y=(3/2)x-1
Sn =(3/2)an-1
n=1, a1= 2
an = Sn -S(n-1)
=(3/2)an -(3/2)a(n-1)
an = 3a(n-1)
=3^(n-1) .a1
=2.3^(n-1)
let
dn = d1+(n-1)d
d1 = an
d(n+2) = a(n+1)

a(n+1) - an = (n+2)d
4.3^(n-1) = (n+2)d
d = 4.3^(n-1) /(n+2)

dn = d1+ (n-1)d
= 2.3^(n-1) +(n-1).4.3^(n-1) /(n+2)
= 2.3^(n-1) + 4.3^(n-1) - 3/[4.3^(n-1)]
d1+d2+...+dn
= 3^n -1 + 2(3^n -1) - (3/4)( 1- 1/3^n )/(1-1/3)
= 3(3^n -1) - ( 1- 1/3^n )