数列{an}满足an>0,其前n项和为Sn满足2Sn=an²+an,则an=
问题描述：
数列{an}满足an>0,其前n项和为Sn满足2Sn=an²+an,则an=
答
取n=1,则2a(1) = 2S(1) =(a(1))^2 +a(1),解得a(1)=1,或a(1)=0(舍去).
2S(n+1)=(a(n+1))^2 + a(n+1)
2S(n)=(a(n))^2 + a(n)
两式相减可得
2a(n+1)=(a(n+1))^2 -(a(n))^2 + a(n+1) - a(n)
所以[a(n+1) + a(n)]*[a(n+1) -a(n)-1]=0.
因为a(n)>0,所以a(n+1)=a(n)+1,所以a(n)=a(1) + n-1=n.