高中数学：数列通项问题？

Question

如题所述

Answer 1

Sn =(4/3)an +2/3
n=1
a1=(4/3)a1 +2/3
(1/3)a1 =-2/3
a1=-2
Sn =(4/3)an +2/3 (1)
for n≥2
S(n-1)=(4/3)a(n-1) +2/3 (2)
(1)-(2)
an =(4/3)an -(4/3)a(n-1)
an = 4a(n-1)
=> {an} 是等比数列, q=4
an =4^(n-1) .a1
=-2. 4^(n-1)本回答被提问者采纳

Answer 2

sn=4/3an+2/3
s[n-1]=4/3a[n-1]+2/3
那sn-s[n-1]=an=4/3an-4/3a[n-1]
an=4a[n-1]
这种题目,注意考察sn-s[n-1]=an