28.已知数列{an}\{a_{n}\}{an},{bn}\{b_{n}\}{bn}的前nnn项和分别为SnS_{n}Sn,TnT_{n}Tn,且b1=12{b_1}=\frac{1}{2}b1=21,Sn=12n2+12n{S_n}=\frac{1}{2}{n^2}+\frac{1}{2}nSn=21n2+21n,当n>1n>1n>1时,满足2bnan−1=bn−1an2b_{n}a_{n-1}=b_{n-1}a_{n}2bnan−1=bn−1an.(1)求ana_{n}an;(2)求TnT_{n}Tn.