已知数列an的前n项为Sn,且a1=4/3,(4^n-1)an=3(4^(n-1))Sn,求Sn的通项式和当bn=n/3an时bn的前n项和Tn

(4^n-1)an=3(4^(n-1))Sn,即（4^n-1)(Sn-S(n-1))=3(4^(n-1))Sn,计算后得Sn=(4^n-1)S(n-1)/(4^(n-1))将S(n-1)类似的展开可得Sn=(4^n-1)/3S1=((4^n-1)/3)(4/3),寝室快熄灯了,后面的题目就没时间写出来了,望谅解!...