(1)取n=1时,a2a1=S2+S1=2a1+a2,① 取n=2时,=2a1+2a2.②由②-①得,a2(a2-a1)=a2.③ 若a2=0,由①知a1=0;若a2≠0,由③知a2-a1=1.④ 由①④解得a1=+1,a2=2+或a1=1-,a2=2-. 综上所述,a1=0,a2=0或a1=+1,a2=+2或a1=1-,a2=2-. (2)当a1>0时,a1=+1,a2=+2. n≥2时,有(2+)an=S2+Sn,(2+)an-1=S2+Sn-1, ∴(1+)an=(2+)an-1,即an=an-1(n≥2), ∴an=a1()n-1=(+1)()n-1.令bn==1-lg2, 故{bn}是递减的等差数列,从而b1>b2>…>b7=lg>lg1=0, n≥8时,bn≤b8=lg<lg1=0,故n=7时,Tn取得最大值,T7=7-lg2 |