数列{an}前n项和为Sn,且an+Sn=-2n-1 证明{an+2}是等比数列

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数列{an}前n项和为Sn,且an+Sn=-2n-1 证明{an+2}是等比数列
数列{an}前n项和为Sn,且an+Sn=-2n-1 证明{an+2}是等比数列

数列{an}前n项和为Sn,且an+Sn=-2n-1 证明{an+2}是等比数列
an+sn=-2n-1,
当n＝1时,a1+s1=-3,则a1=-3/2.
由已知得：sn=-2n-1-an
当n大于或等于2时,则
an=sn-s(n-1）=-2n-1-an-[-2(n-1)-1-a(n-1)]=-2+a(n-1)-an
∴2an=a(n-1)-2,
所以2(an+2)=a(n-1)+2,
即(an+2)/[ a(n-1)+2]=1/2,
∴数列{an+2}为等比数列,且公比为1/2.

证明：Sn-Sn-1=an=-2n-1-an+2n+1+an-1
2an=an-1
an/an-1=2
所以{an}成等比
所以{an+2}成等比

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