Question

已知数列{an}的首项为a1=1,其前n项和为Sn,且对任意正整数n有n,an,Sn成等差数列.

(1)求证:数列{Sn+n+2}成等比数列.

(2)求数列{an}的通项公式.

 

Answer 1

(1)见解析 (2) an=2n-1

【解析】(1)因为n,an,Sn成等差数列,所以2an=Sn+n,由当n≥2时,an=Sn-Sn-1,

所以2(Sn-Sn-1)=Sn+n,

即Sn=2Sn-1+n(n≥2),

所以Sn+n+2=2Sn-1+2n+2=2[Sn-1+(n-1)+2].

又S1+1+2=4≠0,

所以=2,所以数列{Sn+n+2}成等比数列.

(2)由(1)知{Sn+n+2}是以S1+3=a1+3=4为首项,2为公比的等比数列,所以Sn+n+2=4×2n-1=2n+1,又2an=n+Sn,所以2an+2=2n+1,所以an=2n-1.