(1)数列{an}中a1 = 1/3,而且an+1 = Sn+1 – Sn = 1/3n+1 ,所以数列{an}的通项公式an = 1/3n ,n∈N* ,前n项的和Sn = (1/3)(1 – 1/3n)(1 – 1/3) = (1 – 1/3n)/2,即Sn = (1 – 1/3n)/2,n∈N* ;
(2)S1 = a1 = 1/3,a2 ...
(1)数列{an}中a1 = 1/3,而且an+1 = Sn+1 – Sn = 1/3n+1 ,所以数列{an}的通项公式an = 1/3n ,n∈N* ,前n项的和Sn = (1/3)(1 – 1/3n)(1 – 1/3) = (1 – 1/3n)/2,即Sn = (1 – 1/3n)/2,n∈N* ;
(2)S1 = a1 = 1/3,a2 ...