题目

设Sn是数列{an}的前n项和,an>0,且4Sn=an(an+2). (1)求数列{an}的通项公式; (2)设bn=,Tn=b1+b2+…+bn,求证:Tn<. 答案：(1)解4Sn=an(an+2),① 当n=1时,4a1=+2a1,即a1=2. 当n≥2时,4Sn-1=an-1(an-1+2).   ② 由①-②得4an=+2an-2an-1,即2(an+an-1)=(an+an-1)·(an-an-1). ∵an>0,∴an-an-1=2, ∴an=2+2(n-1)=2n. (2)证明∵bn= =, ∴Tn=b1+b2+…+bn=1-+…+1-<设，则（ ）A．y3＞y1＞y2B．y2＞y1＞y3C．y1＞y3＞y2D．y1＞y2＞y3