Question

数列题~~·~~~~~~~~~~~~急啊~~~~~

数列｛an｝ a1=3/2 前n项和Sn且满足2a(n+1)+Sn=3 求an

拜托了~~谢谢

Answer 1

解：2an+1 + Sn =3 Sn=3 - 2an+1 Sn-1=3-2an (n>1)
又an=Sn - Sn-1 =-2an+1 +2an 即an+1/an=1/2 (n>1)
a1=S1=3- 2a2 解得a2=3/4 a2/a1=1/2
∴an+1/an=1/2 (n为正整数)
则｛an｝是首项a1=3/2,公比为1/2的等比数列
∴an=3/2 *(1/2)^(n-1)=3/(2^n)(n为正整数)

Answer 2

a1=3/2
2an+1 + Sn =3 Sn=3 - 2an+1 Sn-1=3-2an (n>1)
又an=Sn - Sn-1 =-2an+1 +2an 即an+1/an=1/2 (n>1)
a1=S1=3- 2a2 解得a2=3/4 a2/a1=1/2
∴an+1/an=1/2 (n为正整数)
则｛an｝是首项a1=3/2,公比为1/2的等比数列
∴an=3/2 *(1/2)^(n-1)=3/(2^n)(n≥N+)