设数列{an}的首项a1=3/2,前n项和为Sn,且满足2a（n+1）+Sn=3（1）求a2及an（2）求满足18/17

Question

设数列{an}的首项a1=3/2,前n项和为Sn,且满足2a（n+1）+Sn=3
（1）求a2及an
（2）求满足18/17

Answer 1

(1)an+1=Sn+1-Sn
2Sn+1-2Sn+Sn=3
2Sn+1=Sn + 2
假设存在k使得
2(Sn+1 +k )=Sn+k
k=-2
所以2(Sn+1 -2 )=Sn - 2
Sn+1-2=1/2*(Sn -2)
令bn=Sn -2,则bn是一首项为b1=S1-2=a1-2=3/2-2=-1/2
公比为1/2的等比数列,bn=-1/2*1/2^(n-1)=-(1/2)^n
b2=1/2*b1
=-1/4=S2-2
=a1+a2-2
a2=2-1/4-a1=1/4
an=Sn - Sn-1=(Sn - 2)-(Sn-1 - 2)=bn-bn-1
=-1/2^n--*1/2^(n-1)=1/2^n,n>1
a1=3/2
(2)Sn=bn+2=2-(1/2)^n
所以18/17