已知数列｛an}的前n项和为sn,且满足(sn+1)-2=sn+an+2n(n为正数）,且 s2=8.求｛an
问题描述：
已知数列｛an}的前n项和为sn,且满足(sn+1)-2=sn+an+2n(n为正数）,且 s2=8.求｛an
}
答
sn+1=sn+an+2n+2sn+1=sn+a(n+1)an+1=an+2n+2s2=8s2=s1+a1+2+2s2=s1+a2a2=a1+4a2-a1=4a2+a1=8a1=2a2=6an+1-an=2(n+1)an-a(n-1)=2n.a2-a1=2*2所以an+1-a1=2*(2+n+1)*n/2=2(n+1)nan+1=2(n+1)n+2所以an=2n(n-1)+2=2n^2-2...