设各项均为正数的数列{an}的前n项和为Sn,已知2a2=a1+a3,数列{根号Sn}是公差为2的等差数列.（1）求数列{an}的通项公式；（2）求数列{an/2^n}的前n项和Tn.

1、√S1=√a1√S2=√(a1+a2)=√a1+2 (1)√S3=√(a1+a2+a3)=√(3a2)=√a1+4 (2)由(1)得a1+a2=a1+4√a1+4√a1=(a2-4)/4代入(2)√(3a2)=(a2-4)/4 +4整理,得a2²-24a2+144=0(a2-12)²=0a2=12a1=(a2-4)²/16=...