Question

已知数列{an}的首项a1=3,前n项和为sn,且sn+1=3sn+2n

是判断数列{an+1}是否成等比数列，并求出数列{an}的通项公式

Answer 1

S(n+1)=3Sn+2n
S(n+1)-Sn=2Sn+2n
a(n+1)=2Sn+2n
an=2S(n-1)+2(n-1) (n>=2)
相减得:
a(n+1)-an=2an+2 (n>=2)
a(n+1)=3an+2 (n>=2)
a(n+1)+1=3[an+1] (n>=2)
从第三项开始,{an+1}是等比关系,后一项等则宏于前一项的3倍.
S(n+1)=3Sn+2n 中,令n=1
S2=3S1+2=11
a2=S2-a1=8
a1+1=4,a2+1=9
a2+1不是a1+1的3倍.
{an+1}不是等比败盯尘数列.
n>=2时,an+1=(a2+1)*3^(n-2)=9*3^(n-2)=3^n
an= 3 (n=1)
an= 3^n-1 (n>=2)
(用大括号括察禅起来.)

Answer 2

S(n+1)=3Sn +2n
S(n+1)+(n+1) +1/2 = 3(Sn +n+1/2)
[S(n+1)+(n+1) +1/2]/早乱(Sn + n+1/2) =3

(Sn + n+1/2)/(S1 + 1+1/2) =3^(n-1)
Sn + n+1/2 = (3/2).3^n
Sn = -n-1/2 +(3/2).3^n
an = Sn -S(n-1)
= 3^n -1
an -1 =3^n
{an+1}是码大等比数列迟睁竖