数列{an}前n项和为Sn,且an+Sn=-2n-1 证明{an+2}是等比数列
问题描述：
数列{an}前n项和为Sn,且an+Sn=-2n-1 证明{an+2}是等比数列
答
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...