解:(1)∵ =t +1, ∴当n=1时,S1=ta2=a1=1, ∴t≠0, 又 +1= +1﹣ , ∴ =t( +1﹣ ), ∴ +1= , ∴当t=﹣1时, +1=0,S1=a1=0, 当t≠﹣1时,数列{ }是等比数列, = , 综上 = . (2)∵Tn=a1+2a2+3a3+…+n ① ∴T1=1,n≥2时又由①可知 +1= ,a2= , ∴ a1+2a3+3a4+…+n +1 ② ①﹣②得 2a2+a3+a4+…+ ﹣n +1 = (a1+a2+a3+…+ )﹣n +1 =﹣1+ ﹣n( +1﹣ ) =﹣1+ ﹣ . Tn=t﹣t +n = . |