各项为正数的数列{an},其前n项和为Sn,且Sn=(√（Sn-1）+√a1)^2(n≥2),数列{bn}的前n项和为Tn,(见下)

题目

各项为正数的数列{an},其前n项和为Sn,且Sn=(√（Sn-1）+√a1)^2(n≥2),数列{bn}的前n项和为Tn,(见下)
若bn=an+1/an+an/an+1,则Tn=?

答案

当n≥2是Sn=(√（Sn-1）+√a1)^2√Sn=√（Sn-1）+√a1√Sn-√（Sn-1）=√a1∴{√Sn}为等差数列,公差为√a1∴√Sn=n√a1∴Sn=n^2 *a1S(n+1)=(n+1)^2*a1a(n+1)=(2n+1)a1an=(2n-1)a1bn=an+1/an+an/an+1=(2n+1)/(2n-1)+(...