Question

设Sn为数列{an}的前n项和,且Sn=3/2(an-1),(n∈N),求数列an的通项公式

n为正整数

Answer 1

解：Sn=3/2(an-1) ①
S(n-1)=3/2(a(n-1)-1) ②
①-②得
Sn-S(n-1)=3/2(an-a(n-1)) ③
即 an=3/2(an-a(n-1)) ④
由④左右移项得 3/2a(n-1)=1/2an
所以 an/a(n-1)=3
又因为 S1=a1
由①得 a1=3/2(a1-1) 得a1=3
所以 an=3^n

Answer 2

A1 = S1 = 3/2(A1-1), A1 = 3
An= Sn - Sn-1 = 3/2(An-An-1),
An = 3An-1
An = 3^n

Bn=4n+3